Rd Sharma Xi 2019 Solutions for Class 12 Science Math Chapter 1 Sets are provided here with simple step-by-step explanations. These solutions for Sets are extremely popular among Class 12 Science students for Math Sets Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma Xi 2019 Book of Class 12 Science Math Chapter 1 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma Xi 2019 Solutions. All Rd Sharma Xi 2019 Solutions for class Class 12 Science Math are prepared by experts and are 100% accurate.

#### Question 1:

Find the equation of the circle with:
(i) Centre (−2, 3) and radius 4.
(ii) Centre (a, b) and radius $\sqrt{{a}^{2}+{b}^{2}}$.
(iii) Centre (0, −1) and radius 1.
(iv) Centre (a cos α, a sin α) and radius a.
(v) Centre (a, a) and radius $\sqrt{2}$ a.

Let (h, k) be the centre of a circle with radius a.
Thus, its equation will be ${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={a}^{2}$.

(i) Here, h = −2, k = 3 and a = 4

∴ Required equation of the circle:
${\left(x+2\right)}^{2}+{\left(y-3\right)}^{2}={4}^{2}$
$⇒{\left(x+2\right)}^{2}+{\left(y-3\right)}^{2}=16$

(ii) Here, h = a, k = b and radius = $\sqrt{{a}^{2}+{b}^{2}}$

∴ Required equation of the circle:
${\left(x-a\right)}^{2}+{\left(y-b\right)}^{2}={a}^{2}+{b}^{2}$
$⇒{x}^{2}+{y}^{2}-2ax-2by=0$

(iii) Here, h = 0, k = −1 and radius = 1

∴ Required equation of the circle:
${\left(x-0\right)}^{2}+{\left(y+1\right)}^{2}={\left(1\right)}^{2}$
$⇒{x}^{2}+{y}^{2}+2y=0$

(iv) Here, h = $a\mathrm{cos}\alpha$, k $a\mathrm{sin}\alpha$ and radius = a

∴ Required equation of the circle:
${\left(x-a\mathrm{cos}\alpha \right)}^{2}+{\left(y-a\mathrm{sin}\alpha \right)}^{2}={\left(a\right)}^{2}$
$⇒{x}^{2}+{a}^{2}{\mathrm{cos}}^{2}\alpha -2ax\mathrm{cos}\alpha +{y}^{2}+{a}^{2}{\mathrm{sin}}^{2}\alpha -2ay\mathrm{sin}\alpha ={a}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{a}^{2}\left({\mathrm{sin}}^{2}\alpha +{\mathrm{cos}}^{2}\alpha \right)-2ax\mathrm{cos}\alpha +{y}^{2}-2ay\mathrm{sin}\alpha ={a}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{a}^{2}-2ax\mathrm{cos}\alpha +{y}^{2}-2ay\mathrm{sin}\alpha ={a}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}-2ax\mathrm{cos}\alpha -2ay\mathrm{sin}\alpha =0$

(v) Here, h = a, k = a and radius = $\sqrt{2}a$

∴ Required equation of the circle:
${\left(x-a\right)}^{2}+{\left(y-a\right)}^{2}={\left(\sqrt{2}a\right)}^{2}$
$⇒{x}^{2}+{a}^{2}-2ax+{y}^{2}+{a}^{2}-2ay=2{a}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}-2ay-2ax=0$

#### Question 2:

Find the centre and radius of each of the following circles:
(i) (x − 1)2 + y2 = 4
(ii) (x + 5)2 + (y + 1)2 = 9
(iii) x2 + y2 − 4x + 6y = 5
(iv) x2 + y2 x + 2y − 3 = 0.

Let (h, k) be the centre of a circle with radius a.
Thus, its equation will be ${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={a}^{2}$.

(i) Given:
(x − 1)2 + y2 = 4

Here, h = 1, k = 0 and a = 2

Thus, the centre is (1, 0) and the radius is 2.

(ii) Given:
(x + 5)2 + (y + 1)2 = 9

Here, h = 5, k = −1 and radius = 3

Thus, the centre is (−5, −1) and the radius is 3.

(iii) Given:
${x}^{2}+{y}^{2}-4x+6y=5$

The given equation can be rewritten as follows:
${\left(x-2\right)}^{2}+{\left(y+3\right)}^{2}-4-9=5$
$⇒{\left(x-2\right)}^{2}+{\left(y+3\right)}^{2}=18$

Thus, the centre is (2, −3).
And, radius = $\sqrt{18}=3\sqrt{2}$

(iv) Given:
${x}^{2}+{y}^{2}-x+2y-3=0$

The given equation can be rewritten as follows:
${\left(x-\frac{1}{2}\right)}^{2}+{\left(y+1\right)}^{2}-\frac{1}{4}-1-3=0$
$⇒{\left(x-\frac{1}{2}\right)}^{2}+{\left(y+1\right)}^{2}=\frac{17}{4}$

Thus, the centre is $\left(\frac{1}{2},-1\right)$  and and the radius is $\frac{\sqrt{17}}{2}$.

#### Question 3:

Find the equation of the circle whose centre is (1, 2) and which passes through the point (4, 6).

Let (h, k) be the centre of a circle with radius a.
Thus, its equation will be ${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={a}^{2}$.

Given:
h = 1, k = 2

∴ Equation of the circle = ${\left(x-1\right)}^{2}+{\left(y-2\right)}^{2}={a}^{2}$     ...(1)

Also, equation (1) passes through (4, 6).

${\left(4-1\right)}^{2}+{\left(6-2\right)}^{2}={a}^{2}$

Substituting the value of a in equation (1):

${\left(x-1\right)}^{2}+{\left(y-2\right)}^{2}=25$

$⇒{x}^{2}+1-2x+{y}^{2}+4-4y=25\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-2x+{y}^{2}-4y=20\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}-2x-4y-20=0$

Thus, the required equation of the circle is ${x}^{2}+{y}^{2}-2x-4y-20=0$.

#### Question 4:

Find the equation of the circle passing through the point of intersection of the lines x + 3y = 0 and 2x − 7y = 0 and whose centre is the point of intersection of the lines x + y + 1 = 0 and x − 2y + 4 = 0.

The point of intersection of the lines x + 3y = 0 and 2x − 7y = 0 is (0, 0).

Let (h, k) be the centre of a circle with radius a.
Thus, its equation will be ${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={a}^{2}$.

The point of intersection of the lines x + y + 1 = 0 and x − 2y + 4 = 0 is (−2, 1).

h = −2, k = 1

∴ Equation of the required circle = ${\left(x+2\right)}^{2}+{\left(y-1\right)}^{2}={a}^{2}$     ...(1)

Also, equation (1) passes through (0, 0).

${\left(0+2\right)}^{2}+{\left(0-1\right)}^{2}={a}^{2}$

Substituting the value of a in equation (1):
${\left(x+2\right)}^{2}+{\left(y-1\right)}^{2}=5$
$⇒{x}^{2}+4+4x+{y}^{2}+1-2y=5\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+4x+{y}^{2}-2y=0$

Hence, the required equation of the circle is ${x}^{2}+{y}^{2}+4x-2y=0$.

#### Question 5:

Find the equation of the circle whose centre lies on the positive direction of y - axis at a distance 6 from the origin and whose radius is 4.

Let (h, k) be the centre of a circle with radius a.
Thus, its equation will be ${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={a}^{2}$.

The centre of the required circle lies on the positive direction of the y-axis at a distance 6 from the origin.
Thus, the coordinates of the centre are (0, 6).

h = 0, k = 6

∴ Equation of the circle = ${\left(x-0\right)}^{2}+{\left(y-6\right)}^{2}={a}^{2}$     ...(1)

Also, a = 4

Substituting the value of a in equation (1):
${\left(x-0\right)}^{2}+{\left(y-6\right)}^{2}=16$

$⇒{x}^{2}+{y}^{2}+36-12y=16\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}-12y+20=0$

Hence, the required equation of the circle is ${x}^{2}+{y}^{2}-12y+20=0$.

#### Question 6:

If the equations of two diameters of a circle are 2x + y = 6 and 3x + 2y = 4 and the radius is 10, find the equation of the circle.

Let (h, k) be the centre of a circle with radius a.
Thus, its equation will be ${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={a}^{2}$.

The intersection point of 2x + y = 6 and 3x + 2y = 4 is (8, −10).

The diameters of a circle intersect at the centre.
Thus, the coordinates of the centre are (8, −10).
h = 8, k = −10

Thus, the equation of the required circle is ${\left(x-8\right)}^{2}+{\left(y+10\right)}^{2}={a}^{2}$     ...(1)

Also, a = 10

Substituting the value of a in equation (1):
${\left(x-8\right)}^{2}+{\left(y+10\right)}^{2}=100$
$⇒{x}^{2}+{y}^{2}-16x+64+100+20y=100\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}-16x+20y+64=0$

Hence, the required equation of the circle is ${x}^{2}+{y}^{2}-16x+20y+64=0$.

#### Question 7:

Find the equation of a circle
(i) which touches both the axes at a distance of 6 units from the origin.
(ii) which touches x-axis at a distance 5 from the origin and radius 6 units.
(iii) which touches both the axes and passes through the point (2, 1).
(iv) passing through the origin, radius 17 and ordinate of the centre is −15.

Let (h, k) be the centre of a circle with radius a.
Thus, its equation will be ${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={a}^{2}$.

(i) Let the required equation of the circle be ${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={a}^{2}$.
It is given that the circle passes through the points (6, 0) and (0, 6).

${\left(6-h\right)}^{2}+{\left(0-k\right)}^{2}={6}^{2}$
And, ${\left(0-h\right)}^{2}+{\left(6-k\right)}^{2}={6}^{2}$

Also, ${h}^{2}+36+{k}^{2}-12k=36$
$⇒{h}^{2}+{k}^{2}=12k$        ...(2)

From (1) and (2), we get:

$12k=12h⇒h=k$

∴ From equation (2), we have:

Consequently, we get:
h = 6

Hence, the required equation of the circle is ${\left(x-6\right)}^{2}+{\left(y-6\right)}^{2}=36$ or ${x}^{2}+{y}^{2}-12x-12y+36=0$.

(ii) Let the required equation of the circle be ${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={a}^{2}$ .
It is given that the circle with radius 6 units touches the x-axis at a distance of 5 units from the origin.
a = 6, h = 5

Hence, the required equation is ${\left(x-5\right)}^{2}+{\left(y-0\right)}^{2}={6}^{2}$ or ${x}^{2}+{y}^{2}-10x-11=0$.

(iii) Let the required equation of the circle be ${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={a}^{2}$.
It is given that the circle touches both the axes.

Thus, the required equation will be ${x}^{2}+{y}^{2}-2ax-2ay+{a}^{2}=0$.

Also, the circle passes through the point (2, 1).

$4+1-4a-2a+{a}^{2}=0$
$⇒{a}^{2}-6a+5=0\phantom{\rule{0ex}{0ex}}⇒{a}^{2}-5a-a+5=0\phantom{\rule{0ex}{0ex}}⇒a=1,5$

Hence, the required equation is ${x}^{2}+{y}^{2}-2x-2y+1=0$ or ${x}^{2}+{y}^{2}-10x-10y+25=0$.

(iv) Let the required equation of the circle be ${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={a}^{2}$.
Given:
k = −15, a = 17

The circle passes through the point (0, 0).
∴ Equation of the circle:
${\left(0-h\right)}^{2}+{\left(0-15\right)}^{2}={\left(17\right)}^{2}$
$h=±8$

Hence, the required equation of the circle is ${\left(x-8\right)}^{2}+{\left(y+15\right)}^{2}={17}^{2}$ or ${\left(x+8\right)}^{2}+{\left(y+15\right)}^{2}={17}^{2}$, i.e. ${x}^{2}+{y}^{2}±16x+30y=0$.

#### Question 8:

Find the equation of the circle which has its centre at the point (3, 4) and touches the straight line 5x + 12y − 1 = 0.

It is given that the centre is at the point (3, 4).
Let the equation of the circle be ${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={a}^{2}$.
∴ Equation of the required circle = ${\left(x-3\right)}^{2}+{\left(y-4\right)}^{2}={a}^{2}$    ...(1)

Also, the circle touches the straight line 5x + 12y − 1 = 0.

So, from equation (1), we have:
${\left(x-3\right)}^{2}+{\left(y-4\right)}^{2}=\frac{3844}{169}$
$⇒{x}^{2}+{y}^{2}-6x-8y=\frac{3844}{169}-25$
$⇒169\left({x}^{2}+{y}^{2}-6x-8y\right)+381=0$

Hence, the required equation of the circle is $169\left({x}^{2}+{y}^{2}-6x-8y\right)+381=0$.

#### Question 9:

Find the equation of the circle which touches the axes and whose centre lies on x − 2y = 3.

If the circle lies in the third quadrant, then its centre will be (−a, −a).

The centre lies on x − 2y = 3.
$-a+2a=3⇒a=3$

∴ Required equation of the circle = ${\left(x+3\right)}^{2}+{\left(y+3\right)}^{2}=9$
=${x}^{2}+{y}^{2}+6x+6y+9=0$

If the circle lies in the fourth quadrant, then its centre will be (a, −a),
$a+2a=3⇒a=1$

∴ Required equation of the circle = ${\left(x-1\right)}^{2}+{\left(y+1\right)}^{2}=1$
= ${x}^{2}+{y}^{2}-2x+2y+1=0$

#### Question 10:

A circle whose centre is the point of intersection of the lines 2x − 3y + 4 = 0 and 3x + 4y − 5 = 0 passes through the origin. Find its equation.

Let the required equation of the circle be ${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={a}^{2}$.
The point of intersection of the lines 2x − 3y + 4 = 0 and 3x + 4y − 5 = 0  is .
∴ Centre = $\left(\frac{-1}{17},\frac{22}{17}\right)$

Also, the circle passes through the origin.

${a}^{2}={\left(\frac{1}{17}\right)}^{2}+{\left(\frac{22}{17}\right)}^{2}=\frac{485}{289}$

Hence, the required equation of the circle is ${\left(x+\frac{1}{17}\right)}^{2}+{\left(y-\frac{22}{17}\right)}^{2}=\frac{485}{289}$.

#### Question 11:

A circle of radius 4 units touches the coordinate axes in the first quadrant. Find the equations of its images with respect to the line mirrors x = 0 and y = 0.

It is given that a circle of radius 4 units touches the coordinate axes in the first quadrant.

Centre of the given circle = (4, 4)

The equation of the given circle is ${\left(x-4\right)}^{2}+{\left(y-4\right)}^{2}=16$.

The images of this circle with respect to the line mirrors x = 0 and y = 0. They have their centres at respectively.

∴ Required equations of the images =  ${\left(x+4\right)}^{2}+{\left(y-4\right)}^{2}=16$ and ${\left(x-4\right)}^{2}+{\left(y+4\right)}^{2}=16$

=  ${x}^{2}+{y}^{2}+8x-8y+16=0$ and ${x}^{2}+{y}^{2}-8x+8y+16=0$

#### Question 12:

Find the equations of the circles touching y-axis at (0, 3) and making an intercept of 8 units on the X-axis.

Case I: The centre lies in first quadrant.

Let the required equation be ${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={a}^{2}$.

Here, AB = 8 units and L (0, 3)

In $△$CAM:
$⇒C{A}^{2}=C{M}^{2}+A{M}^{2}$
$⇒C{A}^{2}={3}^{2}+{4}^{2}\phantom{\rule{0ex}{0ex}}⇒CA=5\phantom{\rule{0ex}{0ex}}⇒CL=CA=5$

∴ Coordinates of the centre =
And, radius of the circle = 5
${\left(x-5\right)}^{2}+{\left(y-3\right)}^{2}=25$, i.e. ${x}^{2}+{y}^{2}-10x-6y=-9$

Case II: The centre lies in the second quadrant.

Coordinates of the centre =
And, radius of the circle= 5
${\left(x+5\right)}^{2}+{\left(y-3\right)}^{2}=25$, i.e. ${x}^{2}+{y}^{2}+10x-6y=-9$

Hence, the equation of the required circle is ${\left(x±5\right)}^{2}+{\left(y-3\right)}^{2}=25$, i.e. ${x}^{2}+{y}^{2}±10x-6y=-9$.

#### Question 13:

Find the equations of the circles passing through two points on Y-axis at distances 3 from the origin and having radius 5.

Let the required equation of the circle be ${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={a}^{2}$.

The circle passes through the points (0, 3) and (0, −3).

${\left(0-h\right)}^{2}+{\left(3-k\right)}^{2}={a}^{2}$       ...(1)
And, ${\left(0-h\right)}^{2}+{\left(-3-k\right)}^{2}={a}^{2}$ ...(2)

Solving (1) and (2), we get:
k=0

Given:
∴ a2 = 25

So, from equation (2), we have:
${h}^{2}+9=25⇒h=±4$

Hence, the required equation is ${\left(x±4\right)}^{2}+{y}^{2}=25$, which can be rewritten as ${x}^{2}±8x+{y}^{2}-9=0$.

#### Question 14:

If the lines 2x 3y = 5 and 3x − 4y = 7 are the diameters of a circle of area 154 square units, then obtain the equation of the circle.

We have Area of circle = 154
$\mathrm{\pi }{r}^{2}=154\phantom{\rule{0ex}{0ex}}⇒{r}^{2}=49$
The intersection of two lines will give us the centre of the circle.
Solving 2x  3y = 5 and 3x − 4y = 7 we get
x = 1 and y = 1
Now, the equation of the circle is given by
${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={r}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-1\right)}^{2}+{\left(y+1\right)}^{2}=49\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}-2x+2y-47=0$

#### Question 15:

If the line y = $\sqrt{3}$x + k touches the circle x2 + y2 = 16, then find the value of k.     [NCERT EXEMPLAR]

The centre and the radius of the circle x2 + y2 = 16 are (0, 0) and 4
Now, the perpendicular distance from the centre of the circle to the tangent y = $\sqrt{3}$x + k is equal to the radius of the circle
$\therefore 4=\left|\frac{\sqrt{3}\left(0\right)-0+k}{\sqrt{{\left(\sqrt{3}\right)}^{2}+{1}^{2}}}\right|\phantom{\rule{0ex}{0ex}}⇒±4=\frac{k}{2}\phantom{\rule{0ex}{0ex}}⇒k=±8$

#### Question 16:

Find the equation of the circle having (1, −2) as its centre and passing through the intersection of the lines 3x + y = 14 and 2x + 5y = 18. [NCERT EXEMPLAR]

Solving 3x + y = 14 and 2x + 5y = 18 we get
x = 4 and y = 2
The radius is equal to the distance between (1, −2) and (4, 2)
$r=\sqrt{{\left(4-1\right)}^{2}+{\left(2+2\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{9+16}\phantom{\rule{0ex}{0ex}}=5$
Now, the equation of the circle is given by
${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={r}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-1\right)}^{2}+{\left(y+2\right)}^{2}=25\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}-2x+4y-20=0$

#### Question 17:

If the lines 3x − 4y + 4 = 0 and 6x − 8y − 7 = 0 are tangents to a circle, then find the radius of  the circle. [NCERT EXEMPLAR]

We have 3x − 4y + 4 = 0 and 6x − 8y − 7 = 0

Since, the slope of both the lines are equal.
Hence, the both the lines are parallel.
The distance between the parralel lines is given by
$\left|\frac{{C}_{1}-{C}_{2}}{\sqrt{{A}^{2}+{B}^{2}}}\right|\phantom{\rule{0ex}{0ex}}=\left|\frac{4+\frac{7}{2}}{\sqrt{{3}^{2}+{4}^{2}}}\right|\phantom{\rule{0ex}{0ex}}=\left|\frac{\frac{15}{2}}{5}\right|\phantom{\rule{0ex}{0ex}}=\frac{3}{2}$
Now, the radius is equal to the half of the distance between the parallel lines(diameter of the circle).
Hence, the radius is given by
$\frac{3}{4}$

#### Question 18:

Show that the point (x, y) given by $x=\frac{2at}{1+{t}^{2}}$ and $y=a\left(\frac{1-{t}^{2}}{1+{t}^{2}}\right)$ lies on a circle for all real values of t such that $-1\le t\le 1$, where a is any given real number. [NCERT EXEMPLAR]

Squaring and adding  $x=\frac{2at}{1+{t}^{2}}$ and $y=a\left(\frac{1-{t}^{2}}{1+{t}^{2}}\right)$, we get
${x}^{2}+{y}^{2}={\left(\frac{2at}{1+{t}^{2}}\right)}^{2}+{a}^{2}{\left(\frac{1-{t}^{2}}{1+{t}^{2}}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}=\frac{4{a}^{2}{t}^{2}+{a}^{2}-2{a}^{2}{t}^{2}+{a}^{2}{t}^{4}}{{\left(1+{t}^{2}\right)}^{2}}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}=\frac{{a}^{2}+2{a}^{2}{t}^{2}+{a}^{2}{t}^{4}}{{\left(1+{t}^{2}\right)}^{2}}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}={a}^{2}\frac{{\left(1+{t}^{2}\right)}^{2}}{{\left(1+{t}^{2}\right)}^{2}}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}={a}^{2}\phantom{\rule{0ex}{0ex}}$
Since, the above equation represents the equation of a circle, hence points (x, y) lies on the circle.

#### Question 19:

The circle x2 + y2 − 2x − 2y + 1 = 0 is rolled along the positive direction of x-axis and makes one complete roll. Find its equation in new-position.

Centre of the given circle =
Radius of the given circle = 1

This circle is rolled along the positive direction of the x-axis. When it makes one complete roll, its centre moves horizontally through a distance equal to its circumference, i.e 2$\mathrm{\pi }$.

Thus, the coordinates of the centre of the new circle will be .

Hence, the required equation of the circle is ${\left(\mathrm{x}-1-2\mathrm{\pi }\right)}^{2}+{\left(y-1\right)}^{2}=1$.

#### Question 20:

One diameter of the circle circumscribing the rectangle ABCD is 4y = x + 7. If the coordinates of A and B are (−3, 4) and (5, 4) respectively, find the equation of the circle.

Clearly, the centre of the circle lies on the line 4y = x + 7.

The circle passes through A (−3, 4) and B (5, 4).

The slope of the segment joining A and B is zero.

Therefore, the slope of the perpendicular bisector of AB is not defined.

Hence, the perpendicular bisector of AB will be parallel to the y-axis and will pass through .

The equation of the perpendicular bisector is $x=1$.

The intersection point of the perpendicular bisector and 4y = x + 7 is .

∴ Centre =

Radius = $\sqrt{{\left(5-1\right)}^{2}+{\left(4-2\right)}^{2}}=\sqrt{20}$

Hence, the required equation of the circle is ${x}^{2}+{y}^{2}-2x-4y-15=0$.

#### Question 21:

If the line 2x y + 1 = 0 touches the circle at the point (2, 5) and the centre of the circle lies on the line x + y − 9 = 0. Find the equation of the circle.

According to question, the centre of the required circle lies on the line x + y − 9 = 0.

Let the coordinates of the centre be .

Let the radius of the circle be a.

Here, a is the distance of the centre from the line 2x y + 1 = 0.

Therefore, the equation of the circle is ${\left(x-t\right)}^{2}+{\left(y-\left(9-t\right)\right)}^{2}={a}^{2}$.        ...(2)

The circle passes through (2, 5).

${\left(2-t\right)}^{2}+{\left(5-\left(9-t\right)\right)}^{2}={a}^{2}$

Substituting t = 6 in (1):

Substituting the values of ${a}^{2}$ and t in equation (2), we find the required equation of circle to be ${\left(x-6\right)}^{2}+{\left(y-3\right)}^{2}=20$.

#### Question 1:

Find the coordinates of the centre and radius of each of the following circles:
(i) x2 + y2 + 6x − 8y − 24 = 0
(ii) 2x2 + 2y2 − 3x + 5y = 7
(iii) 1/2 (x2 + y2) + x cos θ + y sin θ − 4 = 0
(iv) x2 + y2axby = 0

(i) The given equation can be rewritten as ${x}^{2}+{y}^{2}+2\left(3\right)x-2\left(4\right)y-24=0$.

∴ Centre = $\left(-3,-4\right)$
And, radius = $\sqrt{{\left(3\right)}^{2}+{\left(4\right)}^{2}+24}=\sqrt{49}=7$

(ii) The given equation can be rewritten as ${x}^{2}+{y}^{2}-\frac{3x}{2}+\frac{5y}{2}-\frac{7}{2}=0$.

∴ Centre = $\left(\frac{3}{4},\frac{-5}{4}\right)$
And, radius = $\sqrt{{\left(\frac{3}{4}\right)}^{2}+{\left(\frac{-5}{4}\right)}^{2}+\frac{7}{2}}=\sqrt{\frac{34+56}{16}}=\sqrt{\frac{90}{16}}=\frac{3\sqrt{10}}{4}$

(iii) The given equation can be rewritten as ${x}^{2}+{y}^{2}+2x\mathrm{cos}\theta +2y\mathrm{sin}\theta -8=0$.

∴ Centre = $\left(-\mathrm{cos}\theta ,-\mathrm{sin}\theta \right)$
And, radius = $\sqrt{{\left(-\mathrm{cos}\theta \right)}^{2}+{\left(-\mathrm{sin}\theta \right)}^{2}+8}=\sqrt{1+8}=3$

(iv) The given equation can be rewritten as ${x}^{2}+{y}^{2}-\frac{2ax}{2}-\frac{2by}{2}=0$.

∴ Centre = $\left(\frac{a}{2},\frac{b}{2}\right)$
And, radius = $\sqrt{{\left(\frac{a}{2}\right)}^{2}+{\left(\frac{b}{2}\right)}^{2}}=\frac{1}{2}\sqrt{{a}^{2}+{b}^{2}}$

#### Question 2:

Find the equation of the circle passing through the points:
(i) (5, 7), (8, 1) and (1, 3)
(ii) (1, 2), (3, −4) and (5, −6)
(iii) (5, −8), (−2, 9) and (2, 1)
(iv) (0, 0), (−2, 1) and (−3, 2)

(i) Let the required circle be ${x}^{2}+{y}^{2}+2gx+2fy+c=0$.      ...(1)
It passes through (5, 7), (8, 1) and (1, 3).

Substituting the coordinates of these points in equation (1):

$74+10g+14f+c=0$    ...(2)
$65+16g+2f+c=0$      ...(3)
$10+2g+6f+c=0$        ...(4)

Simplifying (2), (3) and (4):

Equation of the required circle:
${x}^{2}+{y}^{2}-\frac{29x}{3}-\frac{19y}{3}+\frac{56}{3}=0$
$⇒$$3\left({x}^{2}+{y}^{2}\right)-29x-19y+56=0$

(ii) Let the required circle be ${x}^{2}+{y}^{2}+2gx+2fy+c=0$.      ...(1)

It passes through (1, 2), (3, −4) and (5, −6).

Substituting the coordinates of these points in equation (1):

$5+2g+4f+c=0$              ...(2)
$25+6g-8f+c=0$            ...(3)
$61+10g-12f+c=0$       ...(4)

Simplifying (2), (3) and (4):

The equation of the required circle is ${x}^{2}+{y}^{2}-22x-4y+25=0$.

(iii) Let the required circle be ${x}^{2}+{y}^{2}+2gx+2fy+c=0$.      ...(1)

It passes through (5, −8), (−2, 9) and (2, 1).

Substituting the coordinates of these points in equation (1):

$89+10g-16f+c=0$    ...(2)
$85-4g+18f+c=0$      ...(3)
$5+4g+2f+c=0$          ...(4)

Simplifying (2), (3) and (4):

The equation of the required circle is ${x}^{2}+{y}^{2}+116x+48y-285=0$.

(iv) Let the required circle be ${x}^{2}+{y}^{2}+2gx+2fy+c=0$.      ...(1)
It passes through (0, 0), (−2, 1) and (−3, 2).

Substituting the coordinates of these points in equation (1):

$c=0$                              ...(2)
$5-4g+2f+c=0$        ...(3)
$13-6g+4f+c=0$     ...(4)

Simplifying (2), (3) and (4):

The equation of the required circle is ${x}^{2}+{y}^{2}-3x-11y=0$.

#### Question 3:

Find the equation of the circle which passes through (3, −2), (−2, 0) and has its centre on the line 2xy = 3.

Let the required equation of the circle be ${x}^{2}+{y}^{2}+2gx+2fy+c=0$.      ...(1)
It is given that the circle passes through (3, −2), (−2, 0).
$13+6g-4f+c=0$   ...(2)
$4-4g+c=0$                 ...(3)

The centre lies on the line 2xy = 3.
$-2g+f-3=0$       ...(4)

Solving (2), (3) and (4):

Hence, the required equation of circle is ${x}^{2}+{y}^{2}+3x+12y+2=0$.

#### Question 4:

Find the equation of the circle which passes through the points (3, 7), (5, 5) and has its centre on the line x − 4y = 1.

Let the required equation of the circle be ${x}^{2}+{y}^{2}+2gx+2fy+c=0$.      ...(1)
It is given that the circle passes through (3, 7), (5, 5).
$58+6g+14f+c=0$   ...(2)
$50+10g+10f+c=0$     ...(3)

The centre lies on the line x − 4y = 1.
$-g+4f-1=0$       ...(4)

Solving (2), (3) and (4):

Hence, the required equation of the circle is ${x}^{2}+{y}^{2}+6x+2y-90=0$.

#### Question 5:

Show that the points (3, −2), (1, 0), (−1, −2) and (1, −4) are concyclic.

Let the required equation of the circle be ${x}^{2}+{y}^{2}+2gx+2fy+c=0$.      ...(1)

It is given that the circle passes through (3, −2), (1, 0), (−1, −2).
$13+6g-4f+c=0$   ...(2)
$1+2g+c=0$                ...(3)
$5-2g-4f+c=0$         ...(4)

Solving (2), (3) and (4):

Theerefore, the equation of the circle is ${x}^{2}+{y}^{2}-2x+4y+1=0$.      ...(5)

We see that the point (1, −4) satisfies the equation (5).

Hence, the points (3, −2), (1, 0), (−1, −2) and (1, −4) are concyclic.

#### Question 6:

Show that the points (5, 5), (6, 4), (−2, 4) and (7, 1) all lie on a circle, and find its equation, centre and radius.

Let the required equation of the circle be ${x}^{2}+{y}^{2}+2gx+2fy+c=0$.      ...(1)

It is given that the circle passes through (5, 5), (6, 4), (−2, 4).
$50+10g+10f+c=0$   ...(2)
$52+12g+8f+c=0$     ...(3)
$20-4g+8f+c=0$       ...(4)

Solving (2), (3) and (4):

Thus, the equation of the circle is ${x}^{2}+{y}^{2}-4x-2y-20=0$.      ...(5)

We see that the point (7, 1) satisfies equation (5).

Hence, the points (5, 5), (6, 4), (−2, 4) and (7, 1) lie on the circle.

Also, centre of the required circle =
Radius of the required circle = $\sqrt{4+1+20}=5$

#### Question 7:

Find the equation of the circle which circumscribes the triangle formed by the lines
(i) x + + 3 = 0, x − y + 1 = 0 and x = 3
(ii) 2x + y − 3 = 0, x + y − 1 = 0 and 3x + 2y − 5 = 0
(iii) x + y = 2, 3x − 4y = 6 and x − y = 0.
(iv) y = x + 2, 3y = 4x and 2y = 3x.

In $∆$ABC:
(i) Let AB represent the line x + + 3 = 0.       ...(1)
Let BC represent the line x − y + 1 = 0.           ...(2)
Let CA represent the line x = 3.                       ...(3)

Intersection point of (1) and (3) is $\left(3,-6\right)$.
Intersection point of (1) and (2) is (−2, −1).
Intersection point of (2) and (3) is (3, 4).

Therefore, the coordinates of A, B and C are $\left(3,-6\right)$, (−2, −1) and (3, 4), respectively.

Let the equation of the circumcircle be ${x}^{2}+{y}^{2}+2gx+2fy+c=0$.
It passes through A, B and C.

∴ $45+6g-12f+c=0$
$5-4g-2f+c=0$
$25+6g+8f+c=0$

Hence, the required equation of the circumcircle is ${x}^{2}+{y}^{2}-6x+2y-15=0$.

(ii) In $∆$ABC:
Let AB represent the line 2x + y − 3 = 0.    ...(1)
Let BC represent the line x + y − 1 = 0.      ...(2)
Let CA represent the line 3x + 2y − 5 = 0.  ...(3)

Intersection point of (1) and (3) is (1, 1).
Intersection point of (1) and (2) is (2, −1).
Intersection point of (2) and (3) is (3, −2).

The coordinates of A, B and C are (1, 1), (2, −1) and (3, −2), respectively.

Let the equation of the circumcircle be ${x}^{2}+{y}^{2}+2gx+2fy+c=0$.
It passes through A, B and C.

∴ $2+2g+2f+c=0$
$5+4g-2f+c=0$
$13+6g-4f+c=0$

Hence, the required equation of the circumcircle is ${x}^{2}+{y}^{2}-13x-5y+16=0$.

(iii) In $∆$ABC:
Let AB represent the line x + y = 2.          ...(1)
Let BC represent the line 3x − 4y = 6.      ...(2)
Let CA represent the line x − y = 0.          ...(3)

Intersection point of (1) and (3) is (1, 1).
Intersection point of (1) and (2) is (2, 0).
Intersection point of (2) and (3) is (−6, −6).

The coordinates of A, B and C are (1, 1), (2, 0) and (−6, −6), respectively.

Let the equation of the circumcircle be ${x}^{2}+{y}^{2}+2gx+2fy+c=0$.
It passes through A, B and C.
∴ $2+2g+2f+c=0$
$4+4g+c=0$
$72-12g-12f+c=0$

Hence, the required equation of the circumcircle is ${x}^{2}+{y}^{2}+4x+6y-12=0$.

(iv)
In $∆$ABC:
(i) Let AB represent the line y = x + 2          ...(1)
Let BC represent the line 3y = 4x                 ...(2)
Let CA represent the line 2y = 3x                 ...(3)

Intersection point of (1) and (3) is (4, 6)
Intersection point of (1) and (2) is (6, 8).
Intersection point of (2) and (3) is (0, 0).

Therefore, the coordinates of A, B and C are (4, 6), (6, 8) and (0, 0) respectively.

Let the equation of the circumcircle be ${x}^{2}+{y}^{2}+2gx+2fy+c=0$.
It passes through A, B and C.

∴ $52+8g+12f+c=0$, $100+12g+16f+c=0$
and

$0++0+0+0+c=0\phantom{\rule{0ex}{0ex}}⇒c=0$

Hence, the required equation of the circumcircle is ${x}^{2}+{y}^{2}-46x+22y=0$.

#### Question 8:

Prove that the centres of the three circles x2 + y2 − 4x − 6y − 12 = 0, x2 + y2 + 2x + 4y − 10 = 0 and x2 + y2 − 10x − 16y − 1 = 0 are collinear.

The given equations of the circles are as follows:
x2 + y2 − 4x − 6y − 12 = 0,   ...(1)
x2 + y2 + 2x + 4y − 10 = 0    ...(2)
And, x2 + y2 − 10x − 16y − 1 = 0    ...(3)

The centre of circle (1) is (2, 3).
The centre of circle (2) is (−1, −2).
The centre of circle (3) is (5, 8).

The area of the triangle formed by the points (2, 3), (−1, −2) and (5, 8) is .
Hence, the centres of the circles x2 + y2 − 4x − 6y − 12 = 0, x2 + y2 + 2x + 4y − 10 = 0 and x2 + y2 − 10x − 16y − 1 = 0 are collinear.

#### Question 9:

Prove that the radii of the circles x2 + y2 = 1, x2 + y2 − 2x − 6y − 6 = 0 and x2 + y2 − 4x − 12y − 9 = 0 are in A.P.

Let the radii of the circles x2 + y2 = 1, x2 + y2 − 2x − 6y − 6 = 0 and x2 + y2 − 4x − 12y − 9 = 0 be , respectively.

Now, ${r}_{2}-{r}_{1}={r}_{3}-{r}_{2}=3$

are in A.P.

#### Question 10:

Find the equation of the circle which passes through the origin and cuts off chords of lengths 4 and 6 on the positive side of the x-axis and y-axis respectively.

According to the question, the circle passes through the origin.
Let the equation of the circle be ${x}^{2}+{y}^{2}-2hx-2ky=0$.

The circle cuts off chords of lengths 4 and 6 on the positive sides of the x-axis and the y-axis, respectively.

∴ Centre =

∴ Required equation:

$⇒{x}^{2}+{y}^{2}-4x-6y=0$

#### Question 11:

Find the equation of the circle concentric with the circle x2 + y2 − 6x + 12y + 15 = 0 and double of its area.

Let the equation of the required circle be ${x}^{2}+{y}^{2}+2gx+2fy+c=0$.
The centre of the circle x2 + y2 − 6x + 12y + 15 = 0 is (3, −6).

Area of the required circle = $2\mathrm{\pi }{r}^{\mathit{2}}$
Here, r = radius of the given circle

Now, r = $\sqrt{9+36-15}=\sqrt{30}$

∴ Area of the required circle = $2\mathrm{\pi }\left(30\right)=60\mathrm{\pi }$

Let R be the radius of the required circle.

$60\mathrm{\pi }=\mathrm{\pi }{R}^{\mathit{2}}\mathit{⇒}{R}^{\mathit{2}}\mathit{=}60$

Thus, the equation of the required circle is ${\left(x-3\right)}^{2}+{\left(y+6\right)}^{2}=60$, i.e. ${x}^{2}+{y}^{2}-6x+12y=15$.

#### Question 12:

Find the equation to the circle which passes through the points (1, 1) (2, 2) and whose radius is 1. Show that there are two such circles.

Let the equation of the required circle be ${x}^{2}+{y}^{2}+2gx+2fy+c=0$.

It passes through (1, 1) and (2, 2).
$2g+2f+c=-2$  ...(1)
And, $4g+4f+c=-8$  ...(2)

From (1) and (2), we have:
$-2g-2f=6⇒g+f=-3$    ...(3)

∴ From (2) and (3), we have:
$c=4$

Using (3), we get:
$g=-2,-1$

Correspondingly, we have:
$f=-1,-2$

Therefore, the required equations of the circles are ${x}^{2}+{y}^{2}-4x-2y+4=0$ and ${x}^{2}+{y}^{2}-2x-4y+4=0$.
Hence, there are two such circles.

#### Question 13:

Find the equation of the circle concentric with x2 + y2 − 4x − 6y − 3 = 0 and which touches the y-axis.

Since, the circles are concentric.
$⇒$Centre of required circle = Centre of x2 + y2 − 4x − 6y − 3 = 0

The centre of the required circle is (2, 3).

We know that if a circle with centre (h, k) touches the y-axis, then h is the radius of the circle.

∴ Equation of the circle:
${\left(x-2\right)}^{2}+{\left(y-3\right)}^{2}={2}^{2}$
$⇒{x}^{2}+{y}^{2}-4x-6y+9=0$

#### Question 14:

If a circle passes through the point (0, 0),(a, 0),(0, b) then find the coordinates of its centre.

The general equation of the circle is x2 + y2 + 2gx + 2fy + c = 0
Now, it is passing through (0, 0)
c = 0
Also, it is passing through (a, 0)
∴  a2 + 2ag = 0
a(a + 2g) = 0
a + 2g = 0
$⇒g=-\frac{a}{2}$
Again, it is passing through (0, b)
b2 + 2bf = 0
b(b + 2f) = 0
b + 2f = 0
$⇒f=-\frac{b}{2}$
The coordinates of its centre are given by

#### Question 15:

Find the equation of the circle which passes through the points (2, 3) and (4,5) and the centre lies on the straight line y − 4x + 3 = 0. [NCERT EXEMPLAR]

The general equation of the circle is x2 + y2 + 2gx + 2fy + c = 0 where the centre of the circle is (−g, −f)
Now, it is passing through (2, 3)
∴ 13 + 4g + 6f + c = 0                       .....(1)
Also, it is passing through (4, 5)
∴  41 + 8g + 10f + c = 0                  .....(2)
g=a2
Now, the centre lies on the straight line y − 4x + 3 = 0
∴ −f + 4g + 3 = 0                 .....(3)
g=a2
Solving (1), (2) and (3), we get
g = −2, f = −5 and c = 25
The equation of the circle is given by x2 + y2 − 4x − 10y + 25 = 0

#### Question 1:

Find the equation of the circle, the end points of whose diameter are (2, −3) and (−2, 4). Find its centre and radius.

(2, −3) and (−2, 4) are the ends points of the diameter of a circle. The equation of this circle is $\left(x-2\right)\left(x+2\right)+\left(y+3\right)\left(y-4\right)=0$.

Equation (1) can be rewritten as
${x}^{2}+{\left(y-\frac{1}{2}\right)}^{2}-\frac{1}{4}-16=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{\left(y-\frac{1}{2}\right)}^{2}=\frac{65}{4}$

∴ Centre is $\left(0,\frac{1}{2}\right)$ and radius is $\frac{\sqrt{65}}{2}$.

#### Question 2:

Find the equation of the circle the end points of whose diameter are the centres of the circles x2 + y2 + 6x − 14y − 1 = 0 and x2 + y2 − 4x + 10y − 2 = 0.

Given:
${x}^{2}+{y}^{2}+6x-14y-1=0$ ...(1)
And, ${x}^{2}+{y}^{2}-4x+10y-2=0$ ...(2)

Equations (1) and (2) can be rewritten as follows:
${\left(x+3\right)}^{2}+{\left(y-7\right)}^{2}=59$
And, ${\left(x-2\right)}^{2}+{\left(y+5\right)}^{2}=31$

Thus, the centres of the circles are (−3, 7) and (2, −5).

Hence, the equation of the circle, the end points of whose diameter are the centres of the given circles, is$\left(x+3\right)\left(x-2\right)+\left(y-7\right)\left(y+5\right)=0$, i.e. ${x}^{2}+{y}^{2}+x-2y-41=0$.

#### Question 3:

The sides of a square are x = 6, x = 9, y = 3 and y = 6. Find the equation of a circle drawn on the diagonal of the square as its diameter.

According to the question:
Sides of the square are x = 6, x = 9, y = 3 and y = 6.

The vertices of the square are (6, 6), (9, 6), (9, 3) and (6, 3).
And, the vertices of two diagonals are (6, 6), (9, 3) and (9, 6), (6, 3).

Hence, the equation of the circle is $\left(x-6\right)\left(x-9\right)+\left(y-6\right)\left(y-3\right)$ or ${x}^{2}+{y}^{2}-15x-9y+72=0$.

#### Question 4:

Find the equation of the circle circumscribing the rectangle whose sides are x − 3y = 4, 3x + y = 22, x − 3y = 14 and 3x + y = 62.

Given:
Sides of the rectangle:
x − 3y = 4     ...(1)
3x + y = 22   ...(2)
x − 3y = 14    ...(3)
And, 3x + y = 62    ...(4)

The intersection of (1) and (2) is (7, 1).
The intersection of (2) and (3) is (8, −2).
The intersection of (3) and (4) is (20, 2).
The intersection of (1) and (4) is (19, 5).

Hence, the vertices of the rectangle are (7, −1), (8, −2), (20, 2) and (19, 5).

The vertices of the diagonals are (7, −1), (20, 2) and (19, 5), (8, −2).

Thus, the required equation of the circle is $\left(x-7\right)\left(x-20\right)+\left(y-1\right)\left(y-2\right)=0$ or ${x}^{2}+{y}^{2}-27x-3y+142=0$.

#### Question 5:

Find the equation of the circle passing through the origin and the points where the line 3x + 4y = 12 meets the axes of coordinates.

Putting x = 0 in 3x + 4y = 12:
y = 3

Putting y = 0 in 3x + 4y = 12:
x = 4

Thus, the line 3x + 4y = 12 meets the axes of coordinates at points A (0, 3) and B (4, 0).

The equation of the circle with AB as the diameter is $\left(x-0\right)\left(x-4\right)+\left(y-3\right)\left(y-0\right)=0$ or ${x}^{2}-4x+{y}^{2}-3y=0$.

Hence, the required equation is ${x}^{2}-4x+{y}^{2}-3y=0$.

#### Question 6:

Find the equation of the circle which passes through the origin and cuts off intercepts a and b respectively from x and y - axes.

Case I:
If the required circle passes through the origin and (a, b), then the end points of the diameter of the circle will be (0, 0) and (a, b).
∴ Required equation of circle:
$\left(x-0\right)\left(x-a\right)+\left(y-0\right)\left(y-b\right)$
or ${x}^{2}+{y}^{2}-ax-by=0$

Case II:
If the required circle passes through the origin and (−a, −b), then the end points of the diameter of the circle will be (0, 0) and (−a, −b).
∴ Required equation of circle:
$\left(x-0\right)\left(x+a\right)+\left(y-0\right)\left(y+b\right)$
or ${x}^{2}+{y}^{2}+ax+by=0$

Hence, the equation of the required circle is ${x}^{2}+{y}^{2}±ax±by=0$.

#### Question 7:

Find the equation of the circle whose diameter is the line segment joining (−4, 3) and (12, −1). Find also the intercept made by it on y-axis.

It is given that the end points of the diameter of the circle are (−4, 3) and (12, −1).
∴ Required equation of circle:
$\left(x+4\right)\left(x-12\right)+\left(y-3\right)\left(y+1\right)$
or ${x}^{2}+{y}^{2}-8x-2y-51=0$   ...(1)

Putting x = 0 in (1):
y2 − 2y − 51 = 0
y2 −  2y − 51 = 0

Hence, the intercepts made by it on the y-axis is $1+2\sqrt{13}-1+2\sqrt{13}=4\sqrt{13}$.

#### Question 8:

The abscissae of the two points A and B are the roots of the equation x2 + 2axb2 = 0 and their ordinates are the roots of the equation x2 + 2pxq2 = 0. Find the equation of the circle with AB as diameter. Also, find its radius.

Roots of equation x2 + 2axb2 = 0 are $-a±\sqrt{{a}^{2}+{b}^{2}}$.
Roots of equation x2 + 2pxq2 = 0 are $-p±\sqrt{{p}^{2}+{q}^{2}}$.

Therefore, coordinates of A and B are respectively.

Hence, equation of circle is .
$⇒{\left(x+a\right)}^{2}-{a}^{2}-{b}^{2}+{\left(y+p\right)}^{2}-{p}^{2}-{q}^{2}=0$
$⇒{x}^{2}+{y}^{2}+2ax+2yp-{p}^{2}-{q}^{2}=0$.

Also, radius of circle is $\sqrt{{a}^{2}+{b}^{2}+{p}^{2}+{q}^{2}}$.

#### Question 9:

ABCD is a square whose side is a; taking AB and AD as axes, prove that the equation of the circle circumscribing the square is x2 + y2a (x + y) = 0.

Given:
ABCD is a square with side a units.

Let AB and AD represent the x-axis and the y-axis, respectively.

Thus, the coordinates of B and D are (a, 0) and (0, a), respectively.

The end points of the diameter of the circle circumscribing the square are B and D.

Thus, equation of the circle circumscribing the square is $\left(x-a\right)\left(x-0\right)+\left(y-0\right)\left(y-a\right)=0$ or ${x}^{2}+{y}^{2}-a\left(x+y\right)=0$.

#### Question 10:

The line 2xy + 6 = 0 meets the circle x2 + y2 − 2y − 9 = 0 at A and B. Find the equation of the circle on AB as diameter.

The equation of the line can be rewritten as $x=\frac{y-6}{2}$.

Substituting the value of x in the equation of the circle, we get:

At y = 0, x = −3
At y = 4, x = −1

Therefore, the coordinates of A and B are .

∴ Equation of the circle with AB as its diameter:
$\left(x+1\right)\left(x+3\right)+\left(y-4\right)\left(y-0\right)=0$
$⇒{x}^{2}+4x+{y}^{2}-4y+3=0$

#### Question 11:

Find the equation of the circle which circumscribes the triangle formed by the lines x = 0, y = 0 and lx + my = 1.

The coordinates of A and B are , respectively.

Here, the end points of the diameter of the circumcircle are A and B.

∴ Required equation of the circle:
$\left(x-0\right)\left(x-\frac{1}{l}\right)+\left(y-\frac{1}{m}\right)\left(y-0\right)=0$
$⇒{x}^{2}-\frac{x}{l}+{y}^{2}-\frac{y}{m}=0$

#### Question 12:

Find the equations of the circles which pass through the origin and cut off equal chords of $\sqrt{2}$ units from the lines y = x and y = − x.

Suppose $a=\sqrt{2}$

From the figure, we see that there will be four circles that pass through the origin and cut off equal chords of length a from the straight lines $y=±x$.

AB, BC, CD and DA are the diameters of the four circles.

Also, ${C}_{1}A=\frac{a}{\sqrt{2}}=O{C}_{1}$

Thus, the coordinates of A are $\left(\frac{a}{\sqrt{2}},\frac{a}{\sqrt{2}}\right)$.

In the same way, we can find the coordinates of B, C and D as $\left(\frac{-a}{\sqrt{2}},\frac{a}{\sqrt{2}}\right),$ $\left(\frac{-a}{\sqrt{2}},\frac{-a}{\sqrt{2}}\right)$ and $\left(\frac{a}{\sqrt{2}},\frac{-a}{\sqrt{2}}\right)$, respectively.

The equation of the circle with AD as the diameter is $\left(x-\frac{a}{\sqrt{2}}\right)\left(x-\frac{a}{\sqrt{2}}\right)+\left(y-\frac{a}{\sqrt{2}}\right)\left(y+\frac{a}{\sqrt{2}}\right)=0$, which can be rewritten as ${x}^{2}+{y}^{2}-\sqrt{2}ax=0$, i.e. ${x}^{2}+{y}^{2}-2x=0$.

Similarly, the equations of the circles with BC, CD and AB as the diameters are ${x}^{2}+{y}^{2}+2x=0$, ${x}^{2}+{y}^{2}+2y=0$ and ${x}^{2}+{y}^{2}-2y=0$, respectively.

#### Question 1:

Write the length of the intercept made by the circle x2 + y2 + 2x − 4y − 5 = 0 on y-axis.

Since the intercept lies on the y-axis, by putting x = 0 in the given equation, we get:

${y}^{2}-4y-5=0$

Thus, the length of the intercept on the y-axis is (5 + 1) = 6 units.

#### Question 2:

Write the coordinates of the centre of the circle passing through (0, 0), (4, 0) and (0, −6).

We need to find the coordinates of the centre of the circle passing through (0, 0), (4, 0) and (0, −6).
Let the equation of the circle be ${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={a}^{2}$.

Putting x = y = 0:
${h}^{2}+{k}^{2}={a}^{2}$        ...(1)

Putting x = 4, y = 0 in the equation of the circle:

Putting x = 0, y = −6 in the equation of the circle:

Hence, the centre of the circle is $\left(2,-3\right)$.

#### Question 3:

Write the area of the circle passing through (−2, 6) and having its centre at (1, 2).

The equation of the required circle is ${\left(x-1\right)}^{2}+{\left(y-2\right)}^{2}={a}^{2}$.
The circle passes through (−2, 6).
${\left(-2-1\right)}^{2}+{\left(6-2\right)}^{2}={a}^{2}$

∴ Area of the required circle =

#### Question 4:

If the abscissae and ordinates of two points P and Q are roots of the equations x2 + 2axb2 = 0 and x2 + 2pxq2 = 0 respectively, then write the equation of the circle with PQ as diameter.

The roots of the equations x2 + 2axb2 = 0 and x2 + 2pxq2 = 0 are $-a±\sqrt{{a}^{2}+{b}^{2}}$ and $-p±\sqrt{{p}^{2}+{q}^{2}}$.
Therefore, the coordinates of P and Q are , respectively.

So, the required equation of the circle is .
$⇒{\left(x+a\right)}^{2}-{a}^{2}-{b}^{2}+{\left(y+p\right)}^{2}-{p}^{2}-{q}^{2}=0$
${x}^{2}+{y}^{2}+2ax+2yp-{p}^{2}-{q}^{2}=0$

#### Question 5:

Write the equation of the unit circle concentric with x2 + y2 − 8x + 4y − 8 = 0.

The centre of the circle x2 + y2 − 8x + 4y − 8 = 0 is (4, −2).

The radius of the unit circle is 1.

∴ Required equation of circle:
${\left(x-4\right)}^{2}+{\left(y+2\right)}^{2}=1$,
$⇒$ ${x}^{2}+{y}^{2}-8x+4y+19=0$

#### Question 6:

If the radius of the circle x2 + y2 + ax + (1 − a) y + 5 = 0 does not exceed 5, write the number of integral values a.

According to the question, we have:

The number of integral values of a is 16.

#### Question 7:

Write the equation of the circle passing through (3, 4) and touching y-axis at the origin.

It is given that the circle touches the y-axis at the origin.

Thus, the centre of the circle is (h,0) and its radius is h.

Hence, the equation of the circle is ${\left(x-h\right)}^{2}+{\left(y\right)}^{2}={h}^{2}$, i.e. ${x}^{2}+{y}^{2}-2xh=0$.

Also, the circle passes through (3, 4).
∴  $25-6h=0⇒h=\frac{25}{6}$

Hence, the required equation of the circle is ${\left(x-\frac{25}{6}\right)}^{2}+{y}^{2}={\left[\frac{25}{6}\right]}^{2}$, i.e. $3\left({x}^{2}+{y}^{2}\right)-25x=0$.

#### Question 8:

If the line y = mx does not intersect the circle (x + 10)2 + (y + 10)2 = 180, then write the set of values taken by m.

Let us put y = mx in the equation (x + 10)2 + (y + 10)2 = 180.

Now, we have:
(x + 10)2 + (mx + 10)2 = 180

On simplifying, we get:

${x}^{2}\left({m}^{2}+1\right)+20x\left(m+1\right)+20=0$

∴ Discriminant (D) = $\sqrt{400{\left(m+1\right)}^{2}-80\left({m}^{2}+1\right)}=4\sqrt{10}\sqrt{\left(2m+1\right)\left(m+2\right)}$

It is given that the line y = mx does not intersect the circle (x + 10)2 + (y + 10)2 = 180.

∴ D <  0

$⇒4\sqrt{10}\sqrt{\left(2m+1\right)\left(m+2\right)}<0\phantom{\rule{0ex}{0ex}}⇒m\in \left(-2,\frac{-1}{2}\right)$

#### Question 9:

Write the coordinates of the centre of the circle inscribed in the square formed by the lines x = 2, x = 6, y = 5 and y = 9.

From the figure, we can see that the vertices of the square are (2, 5), (6, 5), (2, 9) and (6, 9).

The vertices of the diagonals are (2, 9), (6, 5) and (2, 5),(6, 9).

∴ Coordinates of the centre =

#### Question 1:

If the equation of a circle is λx2 + (2λ − 3) y2 − 4x + 6y − 1 = 0, then the coordinates of centre are
(a) (4/3, −1)
(b) (2/3, −1)
(c) (−2/3, 1)
(d) (2/3, 1)

(b)  $\left(\frac{2}{3},-1\right)$

To find the centre:
Coefficient of x2 = Coefficient of y2

Therefore, the given equation can be rewritten as $3{x}^{2}+3{y}^{2}-4x+6y-1=0$.

Thus, the coordinates of the centre is $\left(\frac{2}{3},-1\right)$.

#### Question 2:

If 2x2 + λxy + 2y2 + (λ − 4) x + 6y − 5 = 0 is the equation of a circle, then its radius is
(a) $3\sqrt{2}$
(b) $2\sqrt{3}$
(c) $2\sqrt{2}$
(d) none of these

(d) none of these

The given equation is 2x2 + λxy + 2y2 + (λ − 4) x + 6y − 5 = 0, which can be rewritten as ${x}^{2}+\frac{\lambda xy}{2}+{y}^{2}+\frac{\left(\lambda -4\right)}{2}x+3y-\frac{5}{2}=0$.

Comparing the given equation with ${x}^{2}+{y}^{2}+2gx+2fy+c=0$, we get:
$\lambda =0$

${x}^{2}+{y}^{2}-2x+3y-\frac{5}{2}=0$

∴ Radius = $\sqrt{{\left(-1\right)}^{2}+{\left(\frac{3}{2}\right)}^{2}+\frac{5}{2}}=\sqrt{1+\frac{9}{4}+\frac{5}{2}}=\sqrt{\frac{23}{4}}=\frac{\sqrt{23}}{2}$

#### Question 3:

The equation x2 + y2 + 2x − 4y + 5 = 0 represents
(a) a point
(b) a pair of straight lines
(c) a circle of non-zero radius
(d) none of these

(a) a point

The radius of the given circle = $\sqrt{{1}^{2}+{\left(-2\right)}^{2}-5}=0$

Hence, the radius of the given circle is zero, which represents a point.

#### Question 4:

If the equation (4a − 3) x2 + ay2 + 6x − 2y + 2 = 0 represents a circle, then its centre is
(a) (3, −1)
(b) (3, 1)
(c) (−3, 1)
(d) none of these

(c) (−3, 1)

If the equation (4a − 3) x2 + ay2 + 6x − 2y + 2 = 0 represents a circle, then we have:
Coefficient of x2 = Coefficient of y2
$4a-3=a\phantom{\rule{0ex}{0ex}}$
a = 1

∴ Equation of the circle = ${x}^{2}+{y}^{2}+6x-2y+2=0$

Thus, the coordinates of the centre is .

#### Question 5:

The radius of the circle represented by the equation
3x2 + 3y2 + λxy + 9x + (λ − 6) y + 3 = 0 is
(a) $\frac{3}{2}$

(b) $\frac{\sqrt{17}}{2}$

(c) 2/3

(d) none of these

(a)  $\frac{3}{2}$

The equation of the circle is 3x2 + 3y2 + λxy + 9x + (λ − 6) y + 3 = 0.
∴ Coefficient of xy = 0

$⇒\lambda =0$

Therefore, the radius of the circle is $\sqrt{{\left(\frac{3}{2}\right)}^{2}+{\left(-1\right)}^{2}-1}=\frac{3}{2}$.

#### Question 6:

The number of integral values of λ for which the equation x2 + y2 + λx + (1 − λ) y + 5 = 0 is the equation of a circle whose radius cannot exceed 5, is
(a) 14
(b) 18
(c) 16
(d) none of these

According to the question:

Thus, the number of integral values of $\lambda$ is 16.

#### Question 7:

The equation of the circle passing through the point (1, 1) and having two diameters along the pair of lines x2y2 −2x + 4y − 3 = 0, is
(a) x2 + y2 − 2x − 4y + 4 = 0
(b) x2 + y2 + 2x + 4y − 4 = 0
(c) x2 + y2 − 2x + 4y + 4 = 0
(d) none of these

(a) x2 + y2 − 2x − 4y + 4 = 0

Let the required equation of the circle be ${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={a}^{2}$.

Comparing the given equation x2y2 −2x + 4y − 3 = 0 with $a{x}^{2}+b{y}^{2}+2hxy+2gx+2fy+c=0$, we get:

Intersection point = =

Thus, the centre of the circle is .

The equation of the required circle is ${\left(x-1\right)}^{2}+{\left(y-2\right)}^{2}={a}^{2}$.
Since circle passes through (1, 1), we have:
$1={a}^{2}$

∴ Equation of the required circle:
${\left(x-1\right)}^{2}+{\left(y-2\right)}^{2}=1$
$⇒$ ${x}^{2}+{y}^{2}-2x-4y+4=0$

#### Question 8:

If the centroid of an equilateral triangle is (1, 1) and its one vertex is (−1, 2), then the equation of its circumcircle is
(a) x2 + y2 − 2x − 2y − 3 = 0
(b) x2 + y2 + 2x − 2y − 3 = 0
(c) x2 + y2 + 2x + 2y − 3 = 0
(d) none of these

(a) x2 + y2 − 2x − 2y − 3 = 0

The centre of the circumcircle is (1, 1).

Radius of the circumcircle = $\sqrt{{\left(1+1\right)}^{2}+{\left(1-2\right)}^{2}}=\sqrt{5}$

∴ Equation of the circle:
${\left(x-1\right)}^{2}+{\left(y-1\right)}^{2}=5$
$⇒{x}^{2}+{y}^{2}-2x-2y-3=0$

#### Question 9:

If the point (2, k) lies outside the circles x2 + y2 + x − 2y − 14 = 0 and x2 + y2 = 13 then k lies in the interval
(a) (−3, −2) ∪ (3, 4)
(b) −3, 4
(c) (−∞, −3) ∪ (4, ∞)
(d) (−∞, −2) ∪ (3, ∞)

(c)  (−∞, −3) ∪ (4, ∞)

The given equations of the circles are x2 + y2 + x − 2y − 14 = 0 and x2 + y2 = 13.

Since (2, k) lies outside the given circles, we have:

$4+{k}^{2}+2-2k-14>0$ and $4+{k}^{2}>13$
$⇒{k}^{2}-2k-8>0$ and ${k}^{2}>9$
$⇒\left(k-4\right)\left(k+2\right)>0$ and ${k}^{2}>9$
and

$⇒k\in \left(-\infty ,-3\right)\cup \left(4,\infty \right)$

#### Question 10:

If the point (λ, λ + 1) lies inside the region bounded by the curve $x=\sqrt{25-{y}^{2}}$ and y-axis, then λ belongs to the interval
(a) (−1, 3)
(b) (−4, 3)
(c) (−∞, −4) ∪ (3, ∞)
(d) none of these

(a)  (−1, 3)

The given equation of the curve is ${x}^{2}+{y}^{2}=25$.

Since (λ, λ + 1) lies inside the region bounded by the curve ${x}^{2}+{y}^{2}=25$ and the y-axis, we have:

${\lambda }^{2}+{\left(\lambda +1\right)}^{2}<25$,

#### Question 11:

The equation of the incircle formed by the coordinate axes and the line 4x + 3y = 6 is
(a) x2 + y2 − 6x −6y + 9 = 0
(b) 4 (x2 + y2xy) + 1 = 0
(c) 4 (x2 + y2 + x + y) + 1 = 0
(d) none of these

(b)  4 (x2 + y2xy) + 1 = 0

The line 4x + 3y = 6 cuts the coordinate axes at .

The coordinates of the incentre is .

Here,

Thus, the coordinates of the incentre:

The equation of the incircle:

${\left(x-\frac{1}{2}\right)}^{2}+{\left(y-\frac{1}{2}\right)}^{2}={a}^{2}$

Also, radius of the incircle = $\frac{\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}}{s}$
Here, $s=\frac{a+b+c}{2}=\frac{\frac{5}{2}+\frac{3}{2}+2}{2}=3$

∴ Radius of the incircle = $\frac{\sqrt{3\left(3-a\right)\left(3-b\right)\left(3-c\right)}}{3}$

The equation of circle:
${\left(x-\frac{1}{2}\right)}^{2}+{\left(y-\frac{1}{2}\right)}^{2}=\frac{1}{4}$

#### Question 12:

If the circles x2 + y2 = 9 and x2 + y2 + 8y + c = 0 touch each other, then c is equal to
(a) 15
(b) − 15
(c) 16
(d) − 16

(a)   15
The centre of the circle x2 + y2 = 9 is (0, 0).
Let us denote it by C1.
The centre of the circle x2 + y2+ 8y + c = 0 is (0, −4).
Let us denote it by C2.

The radius of x2 + y2 = 9 is 3 units.

x2 + y2+ 8y + c = 0

Therefore, the radius of the above circle is $\sqrt{16-c}$.

Let the circles touch each other at P.
∴ C1C2 = PC2 + PC1

⇒ PC2 = 4 − 3 = 1
⇒ PC2 = 1 = $\sqrt{16-c}$
c = 15

#### Question 13:

If the circle x2 + y2 + 2ax + 8y + 16 = 0 touches x-axis, then the value of a is
(a) ± 16
(b) ± 4
(c) ± 8
(d) ± 1

(b)   ± 4

The equation of the circle is x2 + y2 + 2ax + 8y + 16 = 0.
Its centre is $\left(-a,-4\right)$ and its radius is a units.

Since the circle touches the x-axis, we have:

$\sqrt{{\left(-a+a\right)}^{2}+{\left(4-0\right)}^{2}}=a$
$a=±4$

#### Question 14:

The equation of a circle with radius 5 and touching both the coordinate axes is
(a) x2 + y2 ± 10x ± 10y + 5 = 0
(b) x2 + y2 ± 10x ± 10y = 0
(c) x2 + y2 ± 10x ± 10y + 25 = 0
(d) x2 + y2 ± 10x ± 10y + 51 = 0

(c)  x2 + y2 ± 10x ± 10y + 25 = 0

Case I: If the circle lies in the first quadrant:

The equation of a circle that touches both the coordinate axes and has radius a is ${x}^{2}+{y}^{2}-2ax-2ay+{a}^{2}=0$.

The given radius of the circle is 5 units, i.e. $a=5$.

Thus, the equation of the circle is ${x}^{2}+{y}^{2}-10x-10y+25=0$.

Case II: If the circle lies in the second quadrant:

The equation of a circle that touches both the coordinate axes and has radius a is ${x}^{2}+{y}^{2}+2ax-2ay+{a}^{2}=0$.

The given radius of the circle is 5 units, i.e. $a=5$.

Thus, the equation of the circle is ${x}^{2}+{y}^{2}+10x-10y+25=0$.

Case III: If the circle lies in the third quadrant:

The equation of a circle that touches both the coordinate axes and has radius a is ${x}^{2}+{y}^{2}+2ax+2ay+{a}^{2}=0$.

The given radius of the circle is 5 units, i.e. $a=5$.

Thus, the equation of the circle is ${x}^{2}+{y}^{2}+10x+10y+25=0$.

Case IV: If the circle lies in the fourth quadrant:

The equation of a circle that touches both the coordinate axes and has radius a is ${x}^{2}+{y}^{2}-2ax+2ay+{a}^{2}=0$.

The given radius of the circle is 5 units, i.e. $a=5$.

Thus, the equation of the circle is ${x}^{2}+{y}^{2}-10x+10y+25=0$.

Hence, the required equation of the circle is x2 + y2 ± 10x ± 10y + 25 = 0.

#### Question 15:

The equation of the circle passing through the origin which cuts off intercept of length 6 and 8 from the axes is
(a) x2 + y2 − 12x − 16y = 0
(b) x2 + y2 + 12x + 16y = 0
(c) x2 + y2 + 6x + 8y = 0
(d) x2 + y2 − 6x − 8y = 0

(d) x2 + y2 − 6x − 8y = 0

The centre of the required circle is .
The radius of the required circle is $\sqrt{{3}^{2}+{4}^{2}}=\sqrt{25}=5$.

Hence, the equation of the circle is as follows:
${\left(x-3\right)}^{2}+{\left(y-4\right)}^{2}={5}^{2}$
$⇒$ ${x}^{2}+{y}^{2}-6x-8y=0$

#### Question 16:

The equation of the circle concentric with x2 + y2 − 3x + 4yc = 0 and passing through (−1, −2) is
(a) x2 + y2 − 3x + 4y − 1 = 0
(b) x2 + y2 − 3x + 4y = 0
(c) x2 + y2 − 3x + 4y + 2 = 0
(d) none of these

(b) x2 + y2 − 3x + 4y = 0

The centre of the circle x2 + y2 − 3x + 4yc = 0 is $\left(\frac{3}{2},-2\right)$.

Therefore, the centre of the required circle is $\left(\frac{3}{2},-2\right)$.
The equation of the circle is ${\left(x-\frac{3}{2}\right)}^{2}+{\left(y+2\right)}^{2}={a}^{2}$.   ...(1)

Also, circle (1) passes through (−1, −2).

$a=\frac{5}{2}$

Substituting the value of a in equation (1):

${\left(x-\frac{3}{2}\right)}^{2}+{\left(y+2\right)}^{2}={\left(\frac{5}{2}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒\frac{{\left(2x-3\right)}^{2}}{4}+{\left(y+2\right)}^{2}=\frac{25}{4}\phantom{\rule{0ex}{0ex}}⇒{\left(2x-3\right)}^{2}+4{\left(y+2\right)}^{2}=25\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}-3x+4y=0$

Hence, the required equation of the circle is ${x}^{2}+{y}^{2}-3x+4y=0$.

#### Question 17:

The circle x2 + y2 + 2gx + 2fy + c = 0 does not intersect x-axis, if
(a) g2 < c
(b) g2 > c
(c) g2 > 2c
(d) none of these

(a) ${g}^{2}

Given:
x2 + y2 + 2gx + 2fy + c = 0   ...(1)

The given circle intersects the x-axis.
The equation of circle becomes x2 + 2gx + c = 0.    ...(2)

Solving equation (2):
∴  Discriminant, D = $\sqrt{4{g}^{2}-4c}\ge 0$
$⇒4{g}^{2}-4c\ge 0\phantom{\rule{0ex}{0ex}}⇒{g}^{2}-c\ge 0\phantom{\rule{0ex}{0ex}}⇒{g}^{2}\ge c$

Hence, if ${g}^{2}, then the given circle will not intersect the x-axis.

#### Question 18:

The area of an equilateral triangle inscribed in the circle x2 + y2 − 6x − 8y − 25 = 0 is
(a) $\frac{225\sqrt{3}}{6}$

(b) 25π

(c) 50π − 100

(d) none of these

(a) $\frac{225\sqrt{3}}{6}$

Let ABC be the required equilateral triangle.
The equation of the circle is x2 + y2 − 6x − 8y − 25 = 0.
Therefore, coordinates of the centre O is .

Radius of the circle = OA = OB = OC = $\sqrt{9+16+25}=5\sqrt{2}$

In $∆$BOD, we have:

$\mathrm{sin}60°=\frac{DB}{BO}\phantom{\rule{0ex}{0ex}}⇒DB=\frac{\sqrt{3}}{2}\left(5\sqrt{2}\right)\phantom{\rule{0ex}{0ex}}⇒BC=2BD=\sqrt{3}\left(5\sqrt{2}\right)=5\sqrt{6}$

Now, area of $△ABC$ = $\frac{\sqrt{3}}{4}B{C}^{2}=\frac{\sqrt{3}}{4}{\left(5\sqrt{6}\right)}^{2}=\frac{\sqrt{3}\left(150\right)}{4}=\frac{\sqrt{3}\left(75\right)}{2}=\frac{\sqrt{3}\left(225\right)}{6}$ square units

#### Question 19:

The equation of the circle which touches the axes of coordinates and the line $\frac{x}{3}+\frac{y}{4}=1$ and whose centres lie in the first quadrant is x2 + y2 − 2cx − 2cy + c2 = 0, where c is equal to
(a) 4
(b) 2
(c) 3
(d) 6

(d) 6

The equation of the circle that touches the axes of coordinates is ${x}^{2}+{y}^{2}-2cx-2cy+{c}^{2}=0$.

Also, ${x}^{2}+{y}^{2}-2cx-2cy+{c}^{2}=0$ touches the line $\frac{x}{3}+\frac{y}{4}=1$ or 4x + 3y $-$12 = 0.
Since the circle lies in the first quadrant, it centre is is .

From the figure, we have:

$\left|\frac{4c+3c-12}{\sqrt{{4}^{2}+{3}^{2}}}\right|=c\phantom{\rule{0ex}{0ex}}⇒\frac{7c-12}{5}=c\phantom{\rule{0ex}{0ex}}⇒c=6$

#### Question 20:

If the circles x2 + y2 = a and x2 + y2 − 6x − 8y + 9 = 0, touch externally, then a =
(a) 1
(b) −1
(c) 21
(d) 16

(a) 1

x2 + y2 = a        ...(1)
And, x2 + y2 − 6x − 8y + 9 = 0     ...(2)

Let circles (1) and (2) touch each other at point P.

The centre of the circle x2 + y2 = a, O, is (0, 0).

The centre of the circle x2 + y2 − 6x − 8y + 9 = 0, C1, is (3, 4).

Also, radius of circle (1) = $\sqrt{a}$ = OP

Radius of circle (2) = $\sqrt{9+16-9}=4$ = C1P

From figure, we have:

${C}_{\mathit{1}}O\mathit{=}{C}_{\mathit{1}}P\mathit{+}OP\phantom{\rule{0ex}{0ex}}\mathit{⇒}\sqrt{{\mathit{3}}^{\mathit{2}}\mathit{+}{\mathit{4}}^{\mathit{2}}}\mathit{=}4+\sqrt{\mathrm{a}}\phantom{\rule{0ex}{0ex}}\mathit{⇒}5\mathit{=}4+\sqrt{\mathrm{a}}\phantom{\rule{0ex}{0ex}}\mathit{⇒}a\mathit{=}1$

#### Question 21:

If (x, 3) and (3, 5) are the extremities of a diameter of a circle with centre at (2, y), then the values of x and y are
(a) (3, 1)
(b) x = 4, y = 1
(c) x = 8, y = 2
(d) none of these

(d) none of these

The end points of the diameter of a circle are (x, 3) and (3, 5).

According to the question, we have:

#### Question 22:

If (−3, 2) lies on the circle x2 + y2 + 2gx + 2fy + c = 0 which is concentric with the circle x2 + y2 + 6x + 8y − 5 = 0, then c =
(a) 11
(b) −11
(c) 24
(d) none of these

(b)   −11

The centre of the circle x2 + y2 + 6x + 8y − 5 = 0 is (−3, −4).

The circle x2 + y2 + 2gx + 2fy + c = 0 is concentric with the circle x2 + y2 + 6x + 8y − 5 = 0.
Thus, the centre of x2 + y2 + 2gx + 2fy + c = 0 is (−3, −4).

Also, it is given that (−3, 2) lies on the circle x2 + y2 + 2gx + 2fy + c = 0.

${\left(-3\right)}^{2}+{2}^{2}+2\left(3\right)\left(-3\right)+2\left(4\right)\left(2\right)+c=0$

$⇒9+4-18+16+c=0\phantom{\rule{0ex}{0ex}}⇒c=-11$

#### Question 23:

Equation of the diameter of the circle x2 + y2 − 2x + 4y = 0 which passes through the origin is
(a) x + 2y = 0
(b) x − 2y = 0
(c) 2x + y = 0
(d) 2xy = 0

(c) 2x + y = 0

Let the diameter of the circle be y = mx.
Since the diameter of the circle passes through its centre, (1, −2) satisfies the equation of the diameter.
$m=-2$

Substituting the value of m in the equation of diameter:
$y=-2x$

Hence, the required equation of the diameter is $2x+y=0$.

#### Question 24:

Equation of the circle through origin which cuts intercepts of length a and b on axes is
(a) x2 + y2 + ax + by = 0
(b) x2 + y2axby = 0
(c) x2 + y2 + bx + ay = 0
(d) none of these

(b) x2 + y2axby = 0

Centre of the circle is $\left(\frac{a}{2},\frac{b}{2}\right)$ and its radius is $\sqrt{{\left(\frac{a}{2}\right)}^{2}+{\left(\frac{b}{2}\right)}^{2}}=\frac{1}{2}\sqrt{{a}^{2}+{b}^{2}}$.

Equation of circle:
${\left(x-\frac{a}{2}\right)}^{2}+{\left(y-\frac{b}{2}\right)}^{2}=\frac{1}{4}\left({a}^{2}+{b}^{2}\right)$
$⇒$${\left(2x-a\right)}^{2}+{\left(2y-b\right)}^{2}=\left({a}^{2}+{b}^{2}\right)$
$⇒$$4{x}^{2}+{a}^{2}-4ax+4{y}^{2}+{b}^{2}-4by={a}^{2}+{b}^{2}$
$⇒$${x}^{2}-ax+{y}^{2}-by=0$

#### Question 25:

If the circles x2 + y2 + 2ax + c = 0 and x2 + y2 + 2by + c = 0 touch each other, then
(a) $\frac{1}{{a}^{2}}+\frac{1}{{b}^{2}}=\frac{1}{c}$

(b) $\frac{1}{{a}^{2}}+\frac{1}{{b}^{2}}=\frac{1}{{c}^{2}}$

(c) a + b = 2c

(d) $\frac{1}{a}+\frac{1}{b}=\frac{2}{c}$

(a) $\frac{1}{{a}^{2}}+\frac{1}{{b}^{2}}=\frac{1}{c}$

Given:
x2 + y2 + 2ax + c = 0    ...(1)
And, x2 + y2 + 2by + c = 0    ...(2)
For circle (1), we have:
Centre = $\left(-a,0\right)$ = C1

For circle (2), we have:
Centre = $\left(0,-b\right)$ = C2

Let the circles intersect at point P.
∴ Coordinates of P = Mid point of C1C2
⇒ Coordinates of P = $\left(\frac{-a+0}{2},\frac{0-b}{2}\right)=\left(\frac{-a}{2},\frac{-b}{2}\right)$

Now, we have:

From (3) and (4), we have:

$\frac{{a}^{2}}{2}={a}^{2}-c\phantom{\rule{0ex}{0ex}}⇒\frac{{a}^{2}}{2}=c\phantom{\rule{0ex}{0ex}}⇒\frac{2}{{a}^{2}}=\frac{1}{c}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{{a}^{2}}+\frac{1}{{a}^{2}}=\frac{1}{c}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{{a}^{2}}+\frac{1}{{b}^{2}}=\frac{1}{c}$

View NCERT Solutions for all chapters of Class 12