RD Sharma 2022 Solutions for Class 10 Maths Chapter 8 - Circles

Answer:

(i) We know that the tangent to a circle is that line which touches the circle at exactly one point. This point at which the tangent touches the circle is known as the ‘point of contact’.

Therefore we have,

The common point of the tangent and the circle is called Point of contact.

(ii) We know that the tangent is perpendicular to the radius of the circle at the point of contact. This also means that the tangent is perpendicular to the diameter of the circle at the point of contact. The diameter of the circle can have at the most one more perpendicular line at the other end where it touches the circle. The perpendicular line at the other end of the diameter is the tangent.

Also we know that two lines which are perpendicular to a common line will be parallel to each other.

Therefore,

A circle may have two parallel tangents.

(iii) From the very basic definition of tangent we know that tangent is a line that intersects the circle at exactly one point. Therefore we have,

A tangent to a circle intersects it in one point.

(iv) From the definition of a secant we know that any line that intersects the circle at 2 points is a secant. Therefore, we have

A line intersecting a circle in two points is called a secant.

(v) One of the properties of the tangent is that it is perpendicular to the radius at the point of contact. Therefore,

The angle between the tangent at the point of contact on a circle and the radius through the point is 90°.

Answer:

We know that circle is made of infinite points which are located at a fixed distance from a particular point. Since at each of these infinite points a tangent can be drawn, we can have infinite number of tangent for a given circle.

Answer:

First let us draw whatever is given in the question. This will help us understand the problem better.

Since the tangent will always be perpendicular to the radius we have drawn OA perpendicular to AB. To find the length of OB we have to use Pythagoras theorem.

Therefore, length of OB is 17 cm.

Answer:

Let us first draw whatever is given so that we can understand the problem better.

Since the tangent to a circle is always perpendicular to the radius of the circle at the point of contact, we have drawn OP perpendicular to PQ. Thus we have a right triangle with one of its sides as the radius. To find the radius we have to use Pythagoras theorem.

Therefore the radius of the circle is 7 cm.

Answer:

Let us put the given data in the form of a diagram.

We have to find TP. From the properties of tangents we know that a tangent will always be at right angles to the radius of the circle at the point of contact. Therefore is a right angle and triangle OTP is a right triangle.

We can find the length of TP using Pythagoras theorem. We have,

Therefore, the length of TP is 15 cm.

Answer:

Let us put the given data in the form of a diagram.

We have to find OT. From the properties of tangents we know that a tangent will always be at right angles to the radius of the circle at the point of contact. Therefore is a right angle and triangle OTP is a right triangle.

We can find the length of TP using Pythagoras theorem. We have,

Therefore, the radius of the circle is 24 cm.

Answer:

Given: PA and PB are the tangents to the circle. 
PA = 12 cm
QC = QD = 3 cm    

To find: PC + PD

PA = PB = 12 cm                (The lengths of tangents drawn from an external point to a circle are equal)
Similarly, QC = AC = 3 cm 
and QD = BD = 3 cm.

Now, PC = PA − AC = 12 − 3 = 9 cm
Similarly, PD = PB − BD = 12 − 3 = 9 cm

Hence, PC + PD = 9 + 9 = 18 cm.

Answer:

Let us first put the given data in the form of a diagram.

We have been asked to prove that the sum of the pair of opposite sides of the quadrilateral is equal to the sum of the other pair.

Therefore, we shall first consider,

AB + DC

But by looking at the figure we have,

AB + DC = AF + FB + DH + HC …… (1)

From the property of tangents we know that the length of two tangents drawn to a circle from a common external point will be equal. Therefore we have the following,

AF = AE

FB = BG

DH = ED

HC = CG

Replacing for all the above in equation (1), we have

AB + DC = AE + BG + ED + CG

AB + DC = (AE + ED) + (BG +CG)

AB + DC = AD + BC

Thus we have proved that the sum of the pair of opposite sides of the quadrilateral is equal to the sum of the other pair.

Answer:

We have been given that.

From the property of tangents we know that a tangent will always be perpendicular to the radius at the point of contact. Therefore,

Looking at the given figure we can rewrite the above equation as follows,

We know that. Therefore,

Now consider. The two sides of this triangle OR and OT are nothing but the radii of the same circle. Therefore,

OR = OT

And hence is an isosceles triangle. We know that the angles opposite to the equal sides of the isosceles triangle will be equal. Therefore,

We have found out that. Therefore,

=

Now consider. We know that sum of all angles of a triangle will always be equal to . Therefore,

Now let us consider the straight line SOT. We know that the angle of a straight line is . Therefore,

From the figure we can see that,

That is,

=

We have found out that. Therefore,

Let us take up now. The sides SO and OR of this triangle are nothing but the radii of the same circle and hence they are equal. Therefore,is an isosceles triangle. In an isosceles triangle, the angles opposite to the two equal sides of the triangle will be equal. Therefore we have,

Also the sum of all angles of a triangle will be equal to. Therefore,

In the previous step we have found out that. Therefore,

Let us now take up . We know from the property of tangents that the angle between the radius of the circle and the tangent at the point of contact will be equal to . Therefore,

=

By looking at the figure we can rewrite the above equation as follows,

In the previous section we have found that. Therefore,

Thus we have found out that .

Answer:

The figure given in the question is below.

From the property of tangents we know that, the length of two tangents drawn from the same external point will be equal. Therefore we have the following,

SA = AP

For our convenience, let us represent SA and AP by a

PB = BQ

Let us represent PB and BQ by b

QC = CR

Let us represent QC and CR by c

DR = DS

Let us represent DR and DS by d

It is given in the problem that,

AB = 6

By looking at the figure we can rewrite the above equation as follows,

AP + PB = 6

a + b = 6

b = 6 − a …… (1)

Similarly we have,

BC = 7

BQ + QC = 7

b + c = 7

Let us substitute the value of b which we have found in equation (1). We have,

6 − a + c = 7

c = a + 1 …… (2)

CD = 4

CR + RD = 4

c + d = 4

Let us substitute the value of c which we have found in equation (2).

a + 1 + d = 4

a + d =3

As per our representations in the previous section, we can write the above equation as follows,

SA + DS = 3

By looking at the figure we have,

AD = 3

Thus we have found that length of side AD is 3 cm.

Answer:

The figure given the question is

From the property of tangents we know that the radius of the circle will always be perpendicular to the tangent at the point of contact. Therefore, OA is perpendicular to AP and triangle OAP is a right triangle. Therefore,

Now consider the smaller circle. Here again the radius OB will be perpendicular to the tangent BP. Therefore, triangle OBP is a right triangle. Hence we have,

Thus we have found that the length of BP is .

Answer:

Let us first consider the quadrilateral OPBQ.

It is given that .

Also from the property of tangents we know that the radius of the circle will always be perpendicular to the tangent at the point of contact. Therefore we have,

We know that sum of all angles of a quadrilateral will always be equal to . Therefore,

Also, in the quadrilateral,

OQ = OP (both are the radii of the same circle)

PB = BQ (from the property of tangents which says that the length of two tangents drawn to a circle from the same external point will be equal)

Since the adjacent sides of the quadrilateral are equal and also since all angles of the quadrilateral are equal to 90°, we can conclude that the quadrilateral OPBQ is a square.
It is given that DS = 5 cm.

From the property of tangents we know that the length of two tangents drawn from the same external point will be equal. Therefore,

DS = DR

DR = 5

It is given that,

AD = 23

DR + RA = 23

5 + RA = 23

RA = 18

Again from the same property of tangents we have,

RA = AQ

We have found out RA = 18. Therefore,

AQ = 18

It is given that AB = 29. That is,

AQ + QB = 29

18 + QB = 29

QB = 11

We have initially proved that OPBQ is a square. QB is one of the sides of the square. Since all sides of the square will be of equal length, we have,

OP = 11

OP is nothing but the radius of the circle.

Thus the radius of the circle is equal to 11 cm.

Answer:

Consider and .

From the property of tangents we know that the length of two tangents drawn form an external point will be equal. Therefore we have,

PA = PB

OB = OA (They are the radii of the same circle)

PO is the common side

Therefore, from SSS postulate of congruency, we have,

Hence,

…… (1)

Now consider and . We have,

(From (1))

PA is the common side.

From the property of tangents we know that the length of two tangents drawn form an external point will be equal. Therefore we have,

PA = PB

From SAS postulate of congruent triangles, we have,

Therefore,

LA = LB

It is given that AB = 16. That is,

LA + LB = 16

LA + LA = 16

2LA = 16

LA = 8

LB = 8

Also, ALB is a straight line. Therefore,

That is,

Since ,

Therefore,

Now let us consider. We have,

Consider. Here,

(Since the radius of the circle will always be perpendicular to the tangent at the point of contact)

Therefore,

…… (1)

Now consider

…… (2)

Since the Left Hand Side of equation (1) is same as the Left Hand Side of equation (2), we can equate the Right Hand Side of the two equations. Hence we have,

= …… (3)

From the figure we can see that,

OP = OL + LP

Therefore, let us replace OP with OL + LP in equation (3). We have,

We have found that OL = 6 and LB = 8. Also it is given that OB = 10. Substituting all these values in the above equation, we get,

Now, let us substitute the value of PL in equation (2). We get,

We know that tangents drawn from an external point will always be equal. Therefore,

PB = PA

Hence, we have,

PA =

Answer:

Let O be the common centre of the two circles and AB be the chord of the larger circle which touches the smaller circle at C.

Join OA and OC. Then,

OC = $\frac{18}{2}$ cm = 9 cm and OA = $\frac{30}{2}$ cm = 15 cm

We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact. Also, the perpendicular drawn from the centre of a circle to a chord bisects the chord.

∴ OC ⊥ AB and C is the mid-point of AB.

In right ∆OCA,

$$\begin{gathered} {\mathrm{OA}}^2={\mathrm{OC}}^2+{\mathrm{AC}}^2 \left(\mathrm{Pythagoras} \mathrm{theorem}\right) \\ \Rightarrow {\mathrm{AC}}^2={\mathrm{OA}}^2-{\mathrm{OC}}^2 \\ \Rightarrow \mathrm{AC}=\sqrt{{15}^2-9^2} \\ \Rightarrow \mathrm{AC}=\sqrt{225-81}=\sqrt{144}=12 \mathrm{cm} \end{gathered}$$

∴ AB = 2AC = 2 × 12 cm = 24 cm

Thus, the required length of the chord is 24 cm.

Answer:

Let the two circles intersect at points X and Y. XY is the common chord.

Suppose A is a point on the common chord and AM and AN be the tangents drawn from A to the circle.

We need to show that AM = AN.

In order to prove the above relation, following property will be used.

“Let PT be a tangent to the circle from an external point P and a secant to the circle through P intersects the circle at points A and B, then PT ² = PA × PB”.

Now, AM is the tangent and AXY is a secant.

∴ AM² = AX × AY ...(1)

AN is the tangent and AXY is a secant.

∴ AN² = AX × AY ...(2)

From (1) and (2), we have

AM² = AN²

∴ AM = AN

Answer:

We know that the lengths of tangents drawn from an external point to a circle are equal.

In the given figure, TQ and TP are tangents drawn to the same circle from an external point T.

∴ TQ = TP               .....(1)

Also, TP and TR are tangents drawn to the same circle from an external point T.

∴ TP = TR               .....(2)

From (1) and (2), we get

TQ = TR

Answer:

The figure given in the question

From the property of tangents we know that the length of two tangents drawn from an external point will we be equal. Hence we have,

PA = PB

PL + LA = PN + NB …… (1)

Again from the same property of tangents we have,

LA = LM (where L is the common external point for tangents LA and LM)

NB = MN (where N is the common external point for tangents NB and MN)

Substituting LM and MN in place of LA and NB in equation (1), we have

PL + LM = PN + MN

Thus we have proved.

Answer:

From the property of tangents we know that the length of two tangents drawn to a circle from the same external point will be equal. Therefore, we have

BQ = BP

Let us denote BP and BQ by x

AP = AR

Let us denote AP and AR by y

RC = QC

Let us denote RC and RQ by z

We have been given that is a right triangle and BC = 6 cm and AB = 8 cm. let us find out AC using Pythagoras theorem. We have,

Consider the perimeter of the given triangle. We have,

AB + BC + AC = 8 + 6 + 10

AB + BC + AC = 24

Looking at the figure, we can rewrite it as,

AP + PB + BQ + QC + AR + RC = 24

Let us replace the sides with the respective x, y and z which we have decided to use.

Now, consider the side AC of the triangle.

AC = 10

Looking at the figure we can say,

AR + RC = 10

y + z = 10 …… (2)

Now let us subtract equation (2) from equation (1). We have,

x + y + z = 12

y + z = 10

After subtracting we get,

x = 2

That is,

BQ = 2, and

BP = 2

Now consider the quadrilateral BPOQ. We have,

BP = BQ (since length of two tangents drawn to a circle from the same external point are equal)

Also,

PO = OQ (radii of the same circle)

It is given that .

From the property of tangents, we know that the tangent will be at right angle to the radius of the circle at the point of contact. Therefore,

We know that sum of all angles of a quadrilateral will be equal to . Therefore,

Since all the angles of the quadrilateral are equal to and the adjacent sides also equal, this quadrilateral is a square. Therefore, all sides will be equal. We have found out that,

BP = 2 cm

Therefore, the radii

PO = 2 cm

Thus the radius of the incircle of the triangle is 2 cm.

Answer:

The given figure is below

(i) The given triangle ABC is a right triangle where side BC is the hypotenuse. Let us now apply Pythagoras theorem. We have,

Looking at the figure we can rewrite the above equation as follows.

…… (1)

From the property of tangents we know that the length of two tangents drawn from the same external point will be equal. Therefore we have the following,

BE = BD

It is given that BD = 30 cm. Therefore,

BE = 30 cm

Similarly,

CD = FC

It is given that CD = 7 cm. Therefore,

FC = 7 cm

Also, on the same lines,

EA = AF

Let us substitute these in equation (1). We get,

Therefore,

AF = 5

Or,

AF = − 42

Since length cannot have a negative value,

AF = 5

(ii) Let us join the point of contact E with the centre of the circle say O. Also, let us join the point of contact F with the centre of the circle O. Now we have a quadrilateral AEOF.

e

In this quadrilateral we have,

(Given in the problem)

(Since the radius will always be perpendicular to the tangent at the point of contact)

(Since the radius will always be perpendicular to the tangent at the point of contact)

We know that the sum of all angles of a quadrilateral will be equal to . Therefore,

Since all the angles of the quadrilateral are equal to 90° and the adjacent sides are equal, this quadrilateral is a square. Therefore all the sides are equal. We have found that

AF = 5

Therefore,

OD = 5

OD is nothing but the radius of the circle.

Thus we have found that AF = 5 cm and radius of the circle is 5 cm.

Answer:

$$\begin{gathered} \angle \mathrm{ACB}=90° \left(\mathrm{Angle} \mathrm{inscribed} \mathrm{in} \mathrm{a} \mathrm{semi}-\mathrm{circle}\right) \\ \angle \mathrm{PCO}=90° \left(\mathrm{PC} \mathrm{is} \mathrm{a} \mathrm{tangent} \mathrm{at} \mathrm{C}\right) \end{gathered}$$
$$\begin{gathered} Now, \angle \mathrm{PCA}=\angle \mathrm{PCO}+\angle \mathrm{OCA} \\ \Rightarrow 110°=90°+\angle \mathrm{OCA} \\ \Rightarrow \angle \mathrm{OCA}=20° \end{gathered}$$
Since, OC = OA           (radii of the circle)
$$\begin{gathered} \angle \mathrm{OCA}=\angle \mathrm{OAC}=20° \\ In ∆ABC, \\ \angle \mathrm{BAC}+\angle \mathrm{ACB}+\angle \mathrm{CBA}=180° \\ \Rightarrow 90°+20°+\angle \mathrm{CBA}=180° \\ \Rightarrow \angle \mathrm{CBA}=70° \end{gathered}$$

Answer:

Given: ∠POR = 130°
So, ∠TSR = $\frac{1}{2}\angle \mathrm{POR}=\frac{1}{2}\times 130°=65°=\angle 2$      .....(1)                   (Since angle subtended by the arc at the centre is double   the angle subtended by it at the remaining part of the circle)
∠POQ = 180º − ∠POR = 180º − 130º = 50º      .....(2)          (Linear pair)       
In $△$POQ,
$$\begin{gathered} \angle 1+\angle \mathrm{POQ}+\angle \mathrm{OQP}=180° \\ \Rightarrow \angle 1+50°+90°=180° \\ \Rightarrow \angle 1=40° \end{gathered}$$
$\mathrm{Now} \angle 1+\angle 2=40°+65°=105°$

Answer:

Here, D, E and F are the points of contact of the circle with the sides BC, AB and AC, respectively.

OD = OE = OF = 4 cm      (Radii of the circle)

We know that the lengths of tangents drawn from an external point to a circle are equal.

∴ BD = BE = 8 cm

CD = CF = 6 cm

AE = AF = x cm (say)

So, BC = BD + CD = 8 cm + 6 cm = 14 cm

AB = AE + BE = x cm + 8 cm = (x + 8) cm

AC = AF + FC = x cm + 6 cm = (x + 6) cm

Also, the tangent at any point of a circle is perpendicular to the radius through the point of contact.

∴ OD ⊥ BC, OE ⊥ AB and OF ⊥ AC

Now,

ar(∆OBC) + ar(∆OAB) + ar(∆OCA) = ar(∆ABC)

$$\begin{gathered} \therefore \frac{1}{2}\times \mathrm{BC}\times \mathrm{OD}+\frac{1}{2}\times \mathrm{AB}\times \mathrm{OE}+\frac{1}{2}\times \mathrm{AC}\times \mathrm{OF}=84 {\mathrm{cm}}^2 \\ \Rightarrow \frac{1}{2}\times 14\times 4+\frac{1}{2}\times \left(x+8\right)\times 4+\frac{1}{2}\times \left(x+6\right)\times 4=84 \\ \Rightarrow 28+2x+16+2x+12=84 \\ \Rightarrow 4x+56=84 \end{gathered}$$
$$\begin{gathered} \Rightarrow 4x=84-56=28 \\ \Rightarrow x=7 \end{gathered}$$

∴ AB = (x + 8) cm = (7 + 8) cm = 15 cm

AC = (x + 6) cm = (7 + 6) cm = 13 cm

Hence, the lengths of sides AB and AC are 15 cm and 13 cm, respectively.

Answer:

It is given that $\angle$AOQ = 58º.

We know that the angle subtended by an arc at the centre is twice the angle subtended by it any point on the remaining part of the circle.

∴ $\angle \mathrm{ABQ}=\frac{1}{2}\angle \mathrm{AOQ}=\frac{1}{2}\times 58°=29°$

Now, AT is the tangent and OA is the radius of the circle through the point of contact A.

$\therefore \angle \mathrm{OAT}=90°$    (Tangent at any point of a circle is perpendicular to the radius through the point of contact)

In ∆ABT,

$\angle \mathrm{BAT}+\angle \mathrm{ABT}+\angle \mathrm{ATB}=180°$   (Angle sum property)

$$\begin{gathered} \Rightarrow 90°+29°+\angle \mathrm{ATB}=180° \left(\angle \mathrm{BAT}=\angle \mathrm{OAT} \mathrm{and} \angle \mathrm{ABT}=\angle \mathrm{ABQ}\right) \\ \Rightarrow \angle \mathrm{ATB}=180°-119°=61° \\ \therefore \angle \mathrm{ATQ}=\angle \mathrm{ATB}=61° \end{gathered}$$

Answer:

In the figure,

. Therefore we can use Pythagoras theorem to find the side PO.

…… (1)

In the problem it is given that,

…… (2)

Substituting this in equation (1), we have,

…… (3)

It is given that the perimeter of is 60 cm. Therefore,

PQ + OQ + PO = 60

Substituting (2) and (3) in the above equation, we have,

Substituting for PQ in equation (2), we have,

OQ is the radius of the circle and QR is the diameter. Therefore,

QR = 2OQ

QR = 30

Substituting for PQ in equation (3), we have,

Thus we have found that PQ = 20 cm, QR = 30 cm and PO = 25 cm.

Answer:

Given that PR = 5 cm.
PR and PQ are the tangents to the inner circle so,
PR = PQ = 5 cm                                (Tangents drawn from an external point to the circle are equal)
Now draw a perpendicular from the centre O to the tangent PS.
PS is the chord of the inner circle. we know that the perpendicular drawn
from the centre of the circle to the chord bisects the chord. So, PQ = QS = 5 cm
PS = PQ + QS = 5 cm + 5 cm = 10 cm

Answer:

Let us first put the given data in the form of a diagram.

From the property of tangents we know that the length of two tangents drawn to a circle from a common external point will always be equal. Therefore,

PA=PB

Consider the triangle PAB. Since we have PA=PB, it is an isosceles triangle. We know that in an isosceles triangle, the angles opposite to the equal sides will be equal. Therefore we have,

Also, sum of all angles of a triangle will be equal to . Therefore,

Since we know that,

Now if we see the values of all the angles of the triangle, all the angles measure. Therefore triangle PAB is an equilateral triangle.

We know that in an equilateral triangle all the sides will be equal.

It is given in the problem that side PA = 10 cm. Therefore, all the sides will measure 10 cm. Hence, AB = 10 cm.

Thus the length of the chord AB is 10 cm.

Answer:

It is given that PA and PB are tangents to the given circle.
$\therefore \angle \mathrm{PAO}=90°$      (Radius is perpendicular to the tangent at the point of contact.)
Now,
$\angle \mathrm{PAB}=50°$          (Given)
$\therefore \angle \mathrm{OAB}=\angle \mathrm{PAO}-\angle \mathrm{PAB}=90°-50°=40°$
In ∆OAB,
OB = OA    (Radii of the circle)
$\therefore \angle \mathrm{OAB}=\angle \mathrm{OBA}=40°$           (Angles opposite to equal sides are equal.)
Now,
$\angle \mathrm{AOB}+\angle \mathrm{OAB}+\angle \mathrm{OBA}=180°$       (Angle sum property)
$\Rightarrow \angle \mathrm{AOB}=180°-40°-40°=100°$

Answer:

Let us first put the given data in the form of a diagram.

From the property of tangents we know that the length of two tangents drawn from the same external point will be equal. Therefore we have,

AR = SA

Let us represent AR and SA by ‘a’.

Similarly,

QB = RB

Let us represent SD and DP by ‘b’

PC = CQ

Let us represent PC and PQ by ‘c’

SD = DP

Let us represent QB and RB by ‘d’

It is given that,

AB = 4

AR + RB =4

a + b = 4

b = 4 − a …… (1)

Similarly,

BC = 5

That is,

b + c = 5

Let us substitute for b from equation (1). We get,

4 − a + c = 5

c − a = 1

c = a + 1 …… (2)

CD = 7

c + d = 7

Let us substitute for c from equation (2). We get,

a + 1 + d = 7

a + d = 6

In the previous section we had represented AS and SR with ‘a’ and SD and DP with ‘b’. We shall now put AS in place of ‘a’ and SD in place of ‘d’. We get,

AS + SD = 6

AD = 6 cm

Therefore, the length of the fourth side of the quadrilateral is 6 cm.

Answer:

Consider the two triangles and .

We have,

is a common angle for both the triangles.

(Given in the problem)

(Since OC is the radius and AC is the tangent to that circle at C and we know that the radius is always perpendicular to the tangent at the point of contact)

Therefore,

From AA similarity postulate we can say that,

~

Since the triangles are similar, all sides of one triangle will be in same proportion to the corresponding sides of the other triangle.

Consider AO′ of and AO of .

Since AO′ and O′X are the radii of the same circle, we have,

AO′ = O′X

Also, since the two circles are equal, the radii of the two circles will be equal. Therefore,

AO′ = XO

Therefore we have

Since ~,

We have found that,

Therefore,

Answer:

The figure given in the question is below

Let us first take up .

We have,

OA = OP (Since they are the radii of the same circle)

Therefore, is an isosceles triangle. From the property of isosceles triangle, we know that, when a median drawn to the unequal side of the triangle will be perpendicular to the unequal side. Therefore,

Now let us take up and .

We know that the radius of the circle will always be perpendicular to the tangent at the point of contact. In this problem, OB is the radius and BC is the tangent and B is the point of contact. Therefore,

Also, from the property of isosceles triangle we have found that

Therefore,

=

is the common angle to both the triangles.

Therefore, from AA postulate of similar triangles,

~

Thus we have proved.

Answer:

In the given figure,

PO = OQ (Since they are the radii of the same circle)

PT = TQ (Length of the tangents from an external point to the circle will be equal) Now considering the angles of the quadrilateral PTQO, we have,

(Given in the problem)

(The radius of the circle will be perpendicular to the tangent at the point of contact)

(The radius of the circle will be perpendicular to the tangent at the point of contact)

We know that the sum of all angles of a quadrilateral will be equal to . Therefore,

Thus we have found that all angles of the quadrilateral are equal to 90°.

Since all angles of the quadrilateral PTQO are equal to 90° and the adjacent sides are equal, this quadrilateral is a square.

We know that in a square, the diagonals will bisect each other at right angles.

Therefore, PQ and OT bisect each other at right angles.

Thus we have proved.

Answer:

In the given figure, let us join D an A.

Consider . We have,

OC = OA (Radii of the same circle)

We know that angles opposite to equal sides of a triangle will be equal. Therefore,

…… (1)

It is clear from the figure that

Now from (1)

Now as BD is tangent therefore

Therefore

From the figure we can see that

Thus we have proved.

Answer:

Disclaimer: The radius of the circle is r.
Given that, two tangents PA and PB are drawn to a circle with centre at O such that OP = 2r.




Applying Pythagoras theorem in $△$AOP,
$$\begin{gathered} {\mathrm{OP}}^2={\mathrm{OA}}^2+{\mathrm{AP}}^2 \\ \Rightarrow {\left(2r\right)}^2=r^2+{\mathrm{AP}}^2 \\ \Rightarrow 4r^2=r^2+{\mathrm{AP}}^2 \\ \Rightarrow 3r^2={\mathrm{AP}}^2 \\ \Rightarrow \mathrm{AP}=\sqrt{3}r\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\left(\because \mathrm{Side} \mathrm{cannot} \mathrm{be} \mathrm{negative}\right) \end{gathered}$$

Since the tangents drawn from an external point to a circle are equal, AP = BP = $\sqrt{3}r$.

In $△$OAP,
$$\begin{gathered} \mathrm{tan\theta }=\frac{r}{\sqrt{3}r}=\frac{1}{\sqrt{3}} \\ \Rightarrow \mathrm{\theta }=30° \end{gathered}$$

Now,
$\alpha =90°-30°=60°$

In $△$OAT,
$$\begin{gathered} \sin \alpha =\sin 60°=\frac{\mathrm{AT}}{r} \\ \Rightarrow \frac{\sqrt{3}}{2}=\frac{\mathrm{AT}}{r} \\ \Rightarrow \mathrm{AT}=\frac{\sqrt{3}r}{2} \\ \Rightarrow \mathrm{BT}=\frac{\sqrt{3}r}{2} \left(\because \mathrm{Line} \mathrm{drawn} \mathrm{from} \mathrm{the} \mathrm{centre} \mathrm{of} \mathrm{the} \mathrm{circle} \mathrm{perpendicular} \mathrm{to} \mathrm{a} \mathrm{chord} \mathrm{bisects} \mathrm{it}\right) \\ \Rightarrow \mathrm{AB}=\sqrt{3}r \end{gathered}$$

Thus, AB = AP = BP = $\sqrt{3}r$. Hence, $△$PAB is equilateral.

Answer:

Let us first put the given data in the form of a diagram.

Consider and. We have,

PO is the common side for both the triangles.

PA = PB(Tangents drawn from an external point will be equal in length)

OB = OA(Radii of the same circle)

Therefore, by SSS postulate of congruency, we have

ΔPOA ΔPOB

Hence,

…… (1)

Now let us consider and . We have,

PL is the common side for both the triangles.

(From equation (1))

PA = PB (Tangents drawn from an external point will be equal in length)

From SAS postulate of congruent triangles,

Therefore,

PL = LB …… (2)

Since AB is a straight line,

Let us now take up ΔOPB. We know that the radius of a circle will always be perpendicular to the tangent at the point of contact. Therefore,

By Pythagoras theorem we have,

It is given that

OP = diameter of the circle

Therefore,

OP = 2OB

Hence,

Consider . We have,

But we have found that,

Also from the figure, we can say

PL = PO − OL 

Therefore,

…… (3)

Also, from , we have

…… (4)

Since Left Hand Sides of equation (3) and equation (4) are same, we can equate the Right Hand Sides of the two equations. Thus we have,

=

We know from the given data, that OP = 2.OB. Let us substitute 2OB in place of PO in the above equation. We get,

Substituting the value of OL and also PO in equation (3), we get,

Also from the figure we get,

AB = PL + LB

From equation (2), we know that PL = LB. Therefore,

AB = 2.LB

We have also found that. We know that tangents drawn from an external point will be equal in length. Therefore, we have

PA = PB

Hence,

PA =

Now, consider . We have,

PA =

PB =

AB =

Since all the sides of the triangle are of equal length, is an equilateral triangle. Thus we have proved.

Answer:

Here, T, S and U are the points of contact of the circle with the sides QR, PQ and PR, respectively.

OT = OS = OU = 8 cm      (Radii of the circle)

We know that the lengths of tangents drawn from an external point to a circle are equal.

∴ QS = QT = 14 cm

RU = RT = 16 cm

PS = PU = x cm (say)

So, QR = QT + TR = 14 cm + 16 cm = 30 cm

PQ = PS + SQ = x cm + 14 cm = (x + 14) cm

PR = PU + UR = x cm + 16 cm = (x + 16) cm

Also, the tangent at any point of a circle is perpendicular to the radius through the point of contact.

∴ OT ⊥ QR, OS ⊥ PQ and OU ⊥ PR

Now,

ar(∆OQR) + ar(∆OPQ) + ar(∆OPR) = ar(∆PQR)

$$\begin{gathered} \therefore \frac{1}{2}\times \mathrm{QR}\times \mathrm{OT}+\frac{1}{2}\times \mathrm{PQ}\times \mathrm{OS}+\frac{1}{2}\times \mathrm{PR}\times \mathrm{OU}=336 {\mathrm{cm}}^2 \\ \Rightarrow \frac{1}{2}\times 30\times 8+\frac{1}{2}\times \left(x+14\right)\times 8+\frac{1}{2}\times \left(x+16\right)\times 8=336 \\ \Rightarrow 120+4x+56+4x+64=336 \\ \Rightarrow 8x+240=336 \end{gathered}$$
$$\begin{gathered} \Rightarrow 8x=336-240=96 \\ \Rightarrow x=12 \end{gathered}$$

∴ PQ = (x + 14) cm = (12 + 14) cm = 26 cm

PR = (x + 16) cm = (12 + 16) cm = 28 cm

Hence, the lengths of sides PQ and PR are 26 cm and 28 cm, respectively.

Answer:

Given data is as follows:

PB = 10 cm

CQ = 2 cm

We have to find the length of PC.

We know that the length of two tangents drawn from the same external point will equal. Therefore,

PB = PA

It is given that PB = 10 cm

Therefore, PA = 10 cm

Also, from the same principle we have,

CQ = CA

It is given that CQ = 2 cm

Therefore, CA = 2cm

From the given figure we can say that,

PC = PA − CA

Now that we know the values of PA and CA, let us substitute the values in the above equation.

PC = 10 − 2

PC = 8 cm

Therefore, length of PC is 8 cm.

Answer:

Consider the quadrilateral OPTQ. It is given that ∠PTQ = 100°.

From the property of the tangent we know that the tangent will always be perpendicular to the radius at the point of contact. Therefore we have,

We know that the sum of all angles of a quadrilateral will always be equal to 360°.

Therefore,

+ ++

Let us substitute the values of all the known angles. We have,

Therefore, the value of angle ∠POQ is 80°.

Answer:

Two parallel tangents can exist at the two ends of the diameter of the circle. Therefore, the distance between the two parallel tangents will be equal to the diameter of the circle. In the problem the radius of the circle is given as 4 cm. Therefore,

Diameter =

Diameter = 8 cm

Hence, the distance between the two parallel tangents is 8 cm.

Answer:

Let us first draw whatever is given for a better understanding of the problem.

Let O be the center of the circle and B be the point of contact.

We know that the radius of the circle will be perpendicular to the tangent at the point of contact. Therefore, we have as a right triangle and we have to apply Pythagoras theorem to find the radius of the triangle.

Therefore, radius of the circle is 3 cm.

Answer:

Two parallel tangents can exist at the two ends of the diameter of the circle. Therefore, the distance between the two parallel tangents will be equal to the diameter of the circle. In the problem the radius of the circle is given as 5 cm. Therefore,

Diameter = 5 × 2

Diameter = 10 cm

Hence, the distance between the two parallel tangents is 10 cm.

Answer:

Here, we have to find the perimeter of triangle PCD.

Perimeter is nothing but sum of all sides of the triangle. Therefore we have,

Perimeter of =

In the given figure we can see that,

=

Therefore,

Perimeter of =

We know that the two tangents drawn to a circle from a common external point will be equal in length. From this property we have,

Now let us replace CQ and QD with CA and DA. We get,

Perimeter of =

Also from the figure we can see that,

Now, let us replace these in the equation for perimeter of. We have,

Perimeter of = PB +PA

Also, from the property of tangents we know that, two tangents drawn to a circle from the same external point will be equal in length. Therefore,

PB = PA

Let us replace PA with PB in the above equation. We get,

Perimeter of = 2PB

It is given in the question that PB = 10 cm. Therefore,

Perimeter of =

Perimeter of = 20 cm

Hence, the perimeter of is 20 cm.

Answer:

CP and CQ are tangents drawn from an external point C to the circle.

∴ CP = CQ (Length of tangents drawn from an external point to the circle are equal)

⇒ CQ = 11 cm (CP = 11 cm)

BQ = CQ − BC = 11 cm − 7 cm = 4 cm

BR and BQ are tangents drawn from an external point B to the circle.

∴ BR = BQ

⇒ BR = 4 cm ( BQ = 4 cm)

Answer:

We are given the following figure

From the figure we get,

BC = BP + PC …… (1)

Now, let us find BP and PC separately.

From the property of tangents we know that when two tangents are drawn to a circle from a common external point, the length of the two tangents from the external point to the respective points of contact will be equal. Therefore we have,

BR = BP

It is given in the problem that BR = 3 cm. Therefore,

BP = 3 cm

Now let us find out PC.

Again using the same property of tangents which says that the length of two tangents drawn to a circle from the same external point will be equal, we have,

PC = QC…… (2)

From the figure we can see that,

QC = AC − AQ…… (3)

Again using the property that length of two tangents drawn to a circle from the same external point will be equal, we have,

AQ = AR

In the problem it is given that,

AR = 4 cm

Therefore,

AQ = 4 cm

Also, the length of AC is also given in the problem.

AC = 11 cm

Let us now substitute the values of AC and AQ in equation (3)

QC = 11 − 4

QC = 7

From equation (2) we can say that,

PC = 7

Finally, let us substitute the values of PC and BP in equation (1)

BC = BP + PC

BC = 3 + 7

BC = 10

Therefore, length of BC is 10 cm.

Answer:

We are given the following figure

From the figure, we have

BC = CQ − BQ…… (1)

Let us now find out the values of CQ and BQ separately.

From the property of tangents we know that length of two tangents drawn from the same external point will be equal. Therefore,

CP = CQ

It is given in the problem that,

CP = 11

Therefore,

CQ = 11

Now let us find out the value of BQ.

Again from the same property of tangents, we know that length of two tangents drawn from the same external point will be equal. Therefore,

BR = BQ

It is given in the problem that,

BR = 4 cm

Therefore,

BQ = 4 cm

Now let us substitute the values of CQ and BQ in equation (1). We have,

BC = 11 − 4

BC = 7

Therefore, length of BC is 7 cm.

Answer:

Let us first draw whatever is given in the problem so that we can understand the problem better.

We have to find the length of AB, which is the chord of the larger circle which touches the smaller circle.

Clearly, OC is the radius of the smaller circle and is touching the tangent AB.

We know that the radius of the circle will always form a right angle with the tangent at the point of contact.

We have draw OA in order to complete the triangle OAC which will be a right triangle.

From the figure it is very clear that OA is the radius of the larger circle which is 5 cm.

We can now find AC using Pythagoras theorem. We have,

Similarly we can find CB. We have,

From the figure we can see that,

AB = AC + CB

Since we have found the values of AC and CB, let us substitute the values in the above equation. We get,

AB = 4 + 4

AB = 8

Therefore, the length of the chord of the larger circle which touches the smaller circle is 8 cm.

Answer:

It is given that tangents PA and PB are drawn from an external point P to a circle with centre O.

∴ PA = PB          (Lengths of tangents drawn from an external point to a circle are equal)

In ∆PAB,

PA = PB

∴ $\angle \mathrm{PBA}=\angle \mathrm{PAB}$     (In a triangle, equal sides have equal angles opposite to them)

Now,

$\angle \mathrm{PAB}+\angle \mathrm{PBA}+\angle \mathrm{APB}=180°$      (Angle sum property)
$$\begin{gathered} \Rightarrow 2\angle \mathrm{PAB}+50°=180° \left(\angle \mathrm{APB}=50°\right) \\ \Rightarrow 2\angle \mathrm{PAB}=180°-50°=130° \\ \Rightarrow \angle \mathrm{PAB}=65° \\ \therefore \angle \mathrm{PBA}=\angle \mathrm{PAB}=65° .....\left(1\right) \end{gathered}$$

Now, PA is the tangent and OA is the radius through the point of contact A.

∴ $\angle \mathrm{OAP}=90°$       (Tangent at any point of a circle is perpendicular to the radius through the point of contact)

Now,

$\angle \mathrm{OAB}=\angle \mathrm{OAP}-\angle \mathrm{PAB}=90°-65°=25° \left[\mathrm{Using}\left(1\right)\right]$

Hence, the measure of $\angle \mathrm{OAB}$ is 25º.

Answer:

Construction: Take any point on major arc PQ and name it S. Join SQ and SP. 
In the given figure, PT is the tangent. So, PT ⊥ PO.
$\angle \mathrm{QPT}=60°$. 
Thus, $\angle \mathrm{OPQ}=90°-60°=30°$
OQ = OP                    (Radii of the circle)
$\angle \mathrm{OQP}=\angle \mathrm{OPQ}=30°$
$$\begin{gathered} \mathrm{In} △\mathrm{OPQ}, \\ \angle \mathrm{OPQ}+\angle \mathrm{OQP}+\angle \mathrm{POQ}=180° \\ \Rightarrow 30°+30°+\angle \mathrm{POQ}=180° \\ \Rightarrow \angle \mathrm{POQ}=120° \end{gathered}$$
Now, $\angle \mathrm{PSQ}=\frac{1}{2}\angle \mathrm{POQ}=\frac{1}{2}\times 120=60°$
PSQR is a cyclic quadrilateral. Thus, 
$$\begin{gathered} \angle \mathrm{PSQ}+\angle \mathrm{PRQ}=180° \\ \Rightarrow 60°+\angle \mathrm{PRQ}=180° \\ \Rightarrow \angle \mathrm{PRQ}=120° \end{gathered}$$

Answer:

In the given figure,
PL and PM are the tangents to the circle with centre O.
$\angle \mathrm{SQL}=50°, \angle \mathrm{SRM}=60°$
$\mathrm{PL}\perp \mathrm{OQ} \mathrm{and} \mathrm{PM}\perp \mathrm{OR}$
Thus, 
$$\begin{gathered} \angle \mathrm{ORS}=90°-60°=30° \\ \angle \mathrm{OQS}=90°-50°=40° \end{gathered}$$
OQ = OS
So, $\angle \mathrm{OQS}=\angle \mathrm{OSQ}=40°$
Similarly. 
OS = OR
So, $\angle \mathrm{ORS}=\angle \mathrm{OSR}=30°$
$$\begin{gathered} \angle \mathrm{RSQ}=\angle \mathrm{QSO}+\angle \mathrm{RSO} \\ \Rightarrow \angle \mathrm{RSQ}=40°+30°=70° \end{gathered}$$

Answer:

In the given figure, PT is the tangent to the circle with centre O. 
$\angle$PBO = 30$°$
OP = OB                  (Radii)
So, $\angle$PBO = $\angle$BPO = 30$°$
$\angle$BPA = 90$°$            (Angle made by the diameter on the arc of the circle)
In âˆ†APB,
$\angle$BPA + $\angle$PBO + $\angle$PAB = 180$°$
⇒ 90$°$ + 30$°$ + $\angle$PAB = 180$°$
⇒ $\angle$PAB = 60$°$
In âˆ†OPA,
OA = OP                 (Radii)
So, $\angle$OPA = OAP = 60$°$
Hence, $\angle$AOP = 180$°$ − 60$°$ − 60$°$ = 60$°$    
In âˆ†OPT,
$\angle$OPT + $\angle$PTO + $\angle$POT = 180$°$
⇒ 90$°$ + $\angle$PTO + 60$°$ = 180$°$
⇒ $\angle$PTO = 30$°$

Answer:

A line intersecting a circle in two distinct points is called a _secant_.

Answer:

AB is the diameter of the circle with centre O. CD and EF are the tangents to the circle at points A and B, respectively.

Now, 

∠OAD = 90º                   (Radius is perpendicular to the tangent at the point of contact)

Also, ∠OBF = 90º          (Radius is perpendicular to the tangent at the point of contact)

∴ ∠OAD + ∠OBF = 90º + 90º = 180º

⇒ CD || EF      (If a pair of consecutive interior angles is supplementary, then the two lines are parallel) 

Hence, the tangents drawn at the end-points of a diameter to a circle are parallel.

The tangents drawn, at the end-points of a diameter, to a circle are __parallel__.

Answer:

PA and PB are two tangents drawn from P to circle with centre O. OP is the line segment joining the point P and centre of the circle.

In âˆ†OAP and âˆ†OBP,

OA = OB      (Radius of the circle)

OP = OP       (Common)

∠OAP = ∠OBP    (Radius is perpendicular to the tangent at the point of contact)

∴ ∆OAP $\cong$ âˆ†OBP      (RHS congruence criterion)

⇒ PA = PB        (CPCT)

Hence, the lengths of two tangents drawn from an external point to a circle are equal.


The lengths of two tangents drawn from an external point to a circle are __equal__.

Answer:

PA and PB are two tangents drawn from P to circle with centre O. OP is the line segment joining the point P and centre of the circle.

In âˆ†OAP and âˆ†OBP,

OA = OB      (Radius of the circle)

OP = OP       (Common)

∠OAP = ∠OBP    (Radius is perpendicular to the tangent at the point of contact)

∴ âˆ†OAP $\cong$ âˆ†OBP      (RHS congruence criterion)

⇒ ∠APO = ∠BPO      (CPCT)

Hence, the two tangents drawn from an external point to a circle are equally inclined to the segment joining the centre to the point.


The two tangents drawn from an external point to a circle are _equally inclined_ to the segment joining the centre to the point.

Answer:

PA and PB are two tangents drawn from P to circle with centre O. OP is the line segment joining the point P and centre of the circle.

In âˆ†OAP and âˆ†OBP,

OA = OB      (Radius of the circle)

OP = OP       (Common)

∠OAP = ∠OBP    (Radius is perpendicular to the tangent at the point of contact)

∴ âˆ†OAP $\cong$ âˆ†OBP      (RHS congruence criterion)

⇒ ∠AOP = ∠BOP      (CPCT)

Hence, the tangents drawn from an external point to a circle subtend equal angles to the centre.
​

Tangents drawn from an external point to a circle subtend __equal__ angles to the centre.

Answer:

ABCD is a parallelogram such that its sides touches the circle with centre O at P, Q, R and S.

We know that the opposite sides of the parallelogram are equal. Therefore,

AB = CD     .....(1)

BC = AD     .....(2)

Also, the length of the tangents drawn from an external point to a circle are equal.

∴ AR = AS      .....(3)

DR = DQ         .....(4)

BP = SB          .....(5) 

CP = CQ         .....(6)

Adding (3), (4), (5) and (6), we get

AR + DR + BP + CP = AS + SB + CQ + DQ

⇒ AD + BC = AB + CD

⇒ 2AD = 2AB          [From (1) and (2)]

⇒ AD = AB       .....(7)

From (1), (2) and (7), we have

AB = BC = CD = AD

Hence, the parallelogram ABCD is a rhombus.


Parallelogram circumscribing a circle is a _rhombus_.

Answer:

Here, AB and CD are common tangents to two circles of unequal radii and having centres O₁ and O₂. 

AB and CD are produced to intersect at P.

Now, PB and PD are tangents drawn from external point P to circle with centre O₂.

∴ PB = PD     .....(1)                (Lengths of tangents drawn from an external point to a circle are equal)

Also, PA and PC are tangents drawn from external point P to circle with centre O₁.

∴ PA = PC      .....(2)               (Lengths of tangents drawn from an external point to a circle are equal)

Subtracting (1) from (2), we get

PA − PB = PC − PD

⇒ AB = CD

If AB and CD are common tangents to two circles of unequal radii, then __AB = CD__.

Answer:

PA and PB are tangents drawn from P to circle with centre O. Let OP and AB intersect at Q.

In âˆ†PAQ and âˆ†PBQ,

AP = BP                   (Length of tangents drawn from an external point to a circle are equal)

∠APQ = ∠BPQ       (Tangents drawn from an external point to a circle are equally inclined to the segment joining the centre to that point) 

PQ = PQ                  (Common)

∴ âˆ†PAQ $\cong$ âˆ†PBQ      (SAS congruence axiom)

⇒ ∠AQP = ∠BQP      (CPCT)

Also,

∠AQP + ∠BQP = 180º       (Linear pair of angles)

⇒ 2∠AQP = 180º               (∠AQP = ∠BQP)

⇒ ∠AQP = 90º 

Therefore, OP is perpendicular to AB i.e. the angle between OP and AB is 90º.

From an external point P, two tangents PA and PB are drawn to the circle with centre O. Then the angle between OP and AB is __90º__.

Answer:

Let PQ and PR be the tangents to the circle with centre O and radius a.

Figure

Since $\angle \mathrm{QPR}=90°$, $\angle \mathrm{QOR}=90°$ by angle sum property of a quadrilateral.
 
Now, OR = OQ = a.       (∵ Radius of same circle)
Since in quadrilateral PQOR pair of adjacent sides are equal and each angle is of 90º.
Thus, PQOR is a square with sides a.

Applying Pythagoras theorem in $△$OQP,
$$\begin{gathered} {\mathrm{OP}}^2={\mathrm{OQ}}^2+{\mathrm{PQ}}^2 \\ \Rightarrow {\mathrm{OP}}^2=a^2+a^2 \\ \Rightarrow {\mathrm{OP}}^2=2a^2 \\ \Rightarrow \mathrm{OP}=\sqrt{2}a\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\left(\because \mathrm{Side} \mathrm{cannot} \mathrm{be} \mathrm{negative}\right) \end{gathered}$$ 

Hence, if the angle between two tangents drawn from a point P to a circle of radius a and centre O is 90°, then OP = $\sqrt{2}a$.

Answer:

AB is a diameter of circle with centre O and AC is the chord of the circle such that ∠BAC = 30°. The tangent at C intersects AB extended at D.

In ∆OAC,

OA = OC    (Radius of circle)

∴ ∠OCA = ∠OAC       (In a triangle, equal sides have equal angles opposite to them)

⇒  ∠OCA = 30°

Now, ∠OCD = 90°    (Radius is perpendicular to the tangent at the point of contact)

∴  ∠ACD  = ∠OCA + ∠OCD = 30° + 90° = 120°

In ∆ACD,

∠A + ∠ACD + ∠D = 180º       (Angle sum property)

⇒ 30° + 120° + ∠D = 180º

⇒ ∠D = 180º − 150° = 30°     .....(1)

Now, ∠ACB = 90°      (Angle in a semi-circle is 90°)

∠ACD = ∠ACB + ∠BCD

⇒ 120° = 90° + ∠BCD

⇒ ∠BCD = 120º − 90° = 30°         .....(2)

In ∆BCD,

∠D = ∠BCD    [From (1) and (2)]

⇒ BC = BD        (In a triangle, equal angles have equal sides opposite to them)

AB is a diameter of a circle and AC is its chord such that ∠BAC = 30°. If the tangent at C intersects AB extended at D, then BC = _  BD _.

Answer:

AB is the chord of a circle with centre O such that ∠AOB = 60º. The tangents at A and B intersect at P.

Now, 

∠OAP = 90º                   (Radius is perpendicular to the tangent at the point of contact)

Also, ∠OBP = 90º          (Radius is perpendicular to the tangent at the point of contact)

In quadrilateral OAPB,

∠AOB + ∠OAP + ∠OBP + ∠APB = 360º         (Angle sum property of quadrilateral)

⇒ 60º + 90º + 90º + ∠APB = 360º

⇒ ∠APB = 360º − 240º = 120º

Thus, the angle between the tangents at A and B is 120º.

If a chord AB subtends an angle of 60° at the centre of a circle, then the angle between the tangents at A and B is __120º__.

Answer:

PT is a tangent drawn from an external point P on the circle with centre O.

∠OTP = 90º         (Radius is perpendicular to the tangent at the point of contact)

So, ∆OPT is a right triangle. OP is the hypotenuse of the right ∆OPT.

∴ OP > PT               (In a right triangle, hypotenuse is the longest side.)

Or PT < OP

Thus, the length of the tangent from an external point P on the circle with centre O is always less than the hypotenuse OP.

The length of the tangent from an external point P on the circle with centre O is always __less than__ OP.

Answer:

PA and PB are tangents drawn from P to circle with centre O.

∠APB = 60º and OA = OB = a  (Radius of the circle)

We know that, the two tangents drawn from an external point to a circle are equally inclined to the segment joining the centre to the point.

∴ ∠APO = ∠BPO = $\frac{60°}{2}$ = 30º

Also, ∠OAP = 90º       (Radius is perpendicular to the tangent at the point of contact)

In right ∆OAP,

$$\begin{gathered} \sin 30°=\frac{\mathrm{OA}}{\mathrm{OP}} \\ \Rightarrow \frac{1}{2}=\frac{a}{\mathrm{OP}} \\ \Rightarrow \mathrm{OP}=2a \end{gathered}$$


If the angle between two tangents drawn from a point P to a circle of radius a and centre O is 60°, then OP = ___2a___.

Answer:

Consider two circles with centres O₁ and O₂ passing through the end points P and Q of a line segment PQ.



We know that the perpendicular bisector of the chord of a circle passes through the centre of the circle.

Thus, the perpendicular bisector PQ passes through the centres O₁ and O₂.

In the same manner, all circles passing through the end points P and Q of a line segment PQ will have their centres lying on the perpendicular bisector of PQ.


If a number of circles pass through the end points P and Q of a line segment PQ, then their centres lie on the perpendicular bisector of ___PQ___.

Answer:

PA and PB are two tangents drawn from P to circle with centre O.

Now, 

∠OAP = 90º                   (Radius is perpendicular to the tangent at the point of contact)

Also, ∠OBP = 90º          (Radius is perpendicular to the tangent at the point of contact)

In quadrilateral OAPB,

∠AOB + ∠OAP + ∠OBP + ∠APB = 360º         (Angle sum property of quadrilateral)

⇒ ∠AOB + 90º + 90º + ∠APB = 360º

⇒ ∠AOB + ∠APB = 360º − 180º = 180º

Thus, the angle between two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segments joining the points of contact at the centre.


The angle between two tangents drawn from an external point to a circle is __supplementary__ to the angle subtended by the line segments joining the points of contact at the centre.

Answer:

PA and PB are two tangents drawn from an external point P to a circle with centre O.

∴ PA = PB      (Lengths of tangents drawn from an external point to a circle are equal)

In ∆APB,

PB = PA        (Proved)

⇒ ∠PAB = ∠PBA       (In a triangle, equal sides have equal angles opposite to them)

⇒ ∠PAB = 65°            (∠PBA = 65°)

Also,

∠PAB + ∠PBA + ∠APB = 180°        (Angle sum property of triangle)

⇒ 65° + 65° + ∠APB = 180°

⇒ ∠APB = 180° − 130° = 50°


PA and PB are two tangents drawn from an external point P to a circle with centre O. If ∠PBA = 65°, then ∠APB = ___50°___.

Answer:

PA and PB are tangents drawn from P to circle with centre O.

∠APB = 80º

We know that, the two tangents drawn from an external point to a circle are equally inclined to the segment joining the centre to the point.

∴ ∠APO = ∠BPO = $\frac{80°}{2}$ = 40º

Also, ∠OAP = 90º       (Radius is perpendicular to the tangent at the point of contact)

In ∆OAP,

∠APO + ∠OAP + ∠POA = 180º         (Angle sum property of triangle)

⇒ 40º + 90º + ∠POA = 180º

⇒ ∠POA = 180º − 130º = 50º


PA and PB are two tangents drawn from an external point P to a circle with centre O. If ∠APB = 80°, then ∠POA = ____50°____.

Answer:

PA and PB are tangents drawn from P to circle with centre O. Let OP and AB intersect at Q.

∠APQ = ∠BPQ = $\frac{60°}{2}$ = 30º       (Tangents drawn from an external point to a circle are equally inclined to the segment joining the centre to that point)

In âˆ†PAQ and âˆ†PBQ,

AP = BP                   (Length of tangents drawn from an external point to a circle are equal)

∠APQ = ∠BPQ       (30º each) 

PQ = PQ                  (Common)

∴ âˆ†PAQ $\cong$ âˆ†PBQ      (SAS congruence axiom)

⇒ ∠AQP = ∠BQP and AQ = BQ      (CPCT)

Also,

∠AQP + ∠BQP = 180º       (Linear pair of angles)

⇒ 2∠AQP = 180º               (∠AQP = ∠BQP)

⇒ ∠AQP = 90º

In right ∆APQ,

$$\begin{gathered} \sin 30°=\frac{\mathrm{AQ}}{\mathrm{AP}} \\ \Rightarrow \frac{1}{2}=\frac{\mathrm{AQ}}{5 \mathrm{cm}} \\ \Rightarrow \mathrm{AQ}=\frac{5}{2}\mathrm{cm} \end{gathered}$$

∴ AB = 2AQ = $2\times \frac{5}{2}$ = 5 cm      (AQ = BQ)

Thus, the length of the chord AB is 5 cm.


If PA and PB are two tangents drawn from an external point P to a circle such that PA = 5 cm and ∠APB = 60°, then the length of chord AB is ___5 cm___.

Answer:

ABCD is a quadrilateral such that its sides touches the circle with centre O at P, Q, R and S.

We know that the lengths of the tangents drawn from an external point to a circle are equal. Therefore,

AR = AS          .....(1)

DR = DQ         .....(2)

BP = SB          .....(3) 

CP = CQ         .....(4)

Adding (1), (2), (3) and (4), we get

AR + DR + BP + CP = AS + SB + CQ + DQ

⇒ AD + BC = AB + CD

⇒ AD + 7 cm = 6 cm + 4 cm

⇒ AD = 10 − 7 = 3 cm


Quadrilateral ABCD is circumscribed to a circle. If AB = 6 cm, BC = 7 cm and CD = 4 cm, then AD = __3 cm__.

Answer:

PT is a tangents drawn from P to circle with centre O.

PT = 10 cm and OT = 5 cm   (Radius of the circle)

Now, 

∠OTP = 90º       (Radius is perpendicular to the tangent at the point of contact)

In right ∆OTP,

$$\begin{gathered} {\mathrm{OP}}^2={\mathrm{OT}}^2+{\mathrm{PT}}^2 \left(\mathrm{Pythagoras} \mathrm{theorem}\right) \\ \Rightarrow {\mathrm{OP}}^2={\left(5\right)}^2+{\left(10\right)}^2 \\ \Rightarrow {\mathrm{OP}}^2=25+100=125 \\ \Rightarrow \mathrm{OP}=5\sqrt{5} \mathrm{cm} \end{gathered}$$

Thus, the distance of the point from the centre of the circle is $5\sqrt{5}$ cm.


The length of the tangent from an external point P to a circle of radius 5 cm is 10 cm. The distance of the point from the centre of the circle is $\underline{ 5\sqrt{5} \mathrm{cm} }$.