# The centre of a circle passing through the points (0,0) & (1,0) & touching the circle x2+y2=9 is: ans is (1/2 , +-root2)..explain AB is a chord of circle x2+y2=25 . The tangents of A & B intersect at C. If (2,3) is the mid-point of AB , the area of quadrilateral OACB is: ams is 50root (3/13) plzz explain 3. Extremities of a dagonal of a rectangle are (0,0) & (4,3).Find the equations of the tangents to the circumcentre ofthe reactangle which are parallel to this diagonal. 4. Find the equ. of circle described on the common cord of the circles x2+y2-4x-5=0 & x2+y2+8y+7=0 as diameter. ans is x2+y2-2x+4y+1=0..explain.

3.... the centre of the circle = (2,3/2).... the slope of diagonal = 3/4 ... so the tangents slope will also be 3/4 .... consider the line y = mx + c ..... m is 3/4 .... u can find c by using the fact dat the perpendicular dist of tangent from centre is equal to radius ...... solve .....

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3.... the centre of the circle = (2,3/2).... the slope of diagonal = 3/4 ... so the tangents slope will also be 3/4 .... consider the line y = mx + c ..... m is 3/4 .... u can find c by using the fact dat the perpendicular dist of tangent from centre is equal to radius ...... solve .....

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3.... the centre of the circle = (2,3/2).... the slope of diagonal = 3/4 ... so the tangents slope will also be 3/4 .... consider the line y = mx + c ..... m is 3/4 .... u can find c by using the fact dat the perpendicular dist of tangent from centre is equal to radius ...... solve .....

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3.... the centre of the circle = (2,3/2).... the slope of diagonal = 3/4 ... so the tangents slope will also be 3/4 .... consider the line y = mx + c ..... m is 3/4 .... u can find c by using the fact dat the perpendicular dist of tangent from centre is equal to radius ...... solve .....

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I m sorry for it being posted so many times .... I m not able to understand anything here .... so juss my mistake ,... I  m sorry

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General Equation of Circle is given as
x2+y2+2gx+2fy+c=0 ---(1)
Since, Circle passes through, (0,0) and (1,0)
Therefore Equation (1) reduces to,
c=0 ---(2)
1+2g+c=0 ---(3),
From (2) and (3), g= -1/2
Also, From As per Question,
Cicle in equation, (1) touches, another circle having equation,
x2+y2=(3)2 ---(4)
SInce the center (0,0) of circle (4) lies on the loci of the circle (1), therefore, The radius of the circle (4) must be equal to diameter of circle(1)
hence, 2√g2+f2-c= 3
=> √1/4+f2=3/2
=> f2= 9/4-1/4
=> f2= 2
=> f= ±√2
Hence Centres of Circle (1) is (-g,-f) or (1/2,±√2)
Hope this helps you!!
Cheers@@ Keep Smiling!!.
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Answer 4 Let the equations of circles be,

S1=x2+y2-4x-5 and S2= x2+y2+8y+7,

Now, Equation of Common Chord will be given as

S1-S2=0

=> -4x-8y-12=0

=> x+2y+3=0

Now, Equation of circle through two circles is

S1+k(S2)=0

=> x2+y2-4x-5+ k(x2+y2+8y+7)=0

=> x2(1+k) +y2(1+k)+ 8ky-4x+7k-5=0

=> x2+y2+ 2(4k)y/(1+k)- 2.2x(1+k)+7k-5/(1+k)=0

Comparing it with general form,

x2+y2+2gx+2fy+c=0

Now, Centre of Circle, (-g,-f)=> (2/1+k, -4k/1+k)

Now, Centre lies on the chord/line,

x+2y+3=0

=> 2/1+k-8k/1+k+3(1+k)/1+k=0

=> 2-8k+3k+3=0

=> k=1

Hence, Required Equation of Circle is,

x2+y2-x+4y+1=0

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Answer 3: Let the extremities of rectangle be A(0,0) and B(4,3)

SInce it is the diameter of the circle, Therefore, |AB|= 5 units, Therefore Radius of circle,

R=5/2 units,

Now, Using mid-point formula, Centre O of circle, is (2, 3/2)

Also, Using two point form , Equation of diagonal is,

y=3/4x

Therefore, m= 3/4, using, y=mx+c

But, Tangent is parallel to above diagonal,

Hence, COnsider a equation of tangent line, y=mx+c

=> y=(3/4)x+c

=> -3x+4y-c=0

The distance of centre to the tangent is equal to radius of circle, i.e. 5/2 units,

Hence, By,

|Ax1+By1+c|/√a2+b2

|-6+6-c|/5=5/2

=> |-c|= 25/2

=> c= 25/2,

Hence Equation of tangent,

-3x+4y+25/2=0

=> 6x-8y-25=0

=> 6x-8y=25

Which is the required equation of tangent.

Hope it helps you!!

CHeers!!@@Keep Smiling!!

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General Equation of Circle is given as x2+y2+2gx+2fy+c=0 ---(1) Since, Circle passes through, (0,0) and (1,0) Therefore Equation (1) reduces to, c=0 ---(2) 1+2g+c=0 ---(3), From (2) and (3), g= -1/2 Also, From As per Question, Cicle in equation, (1) touches, another circle having equation, x2+y2=(3)2 ---(4) SInce the center (0,0) of circle (4) lies on the loci of the circle (1), therefore, The radius of the circle (4) must be equal to diameter of circle(1) hence, 2√g2+f2-c= 3 => √1/4+f2=3/2 => f2= 9/4-1/4 => f2= 2 => f= +-√2 Hence Centres of Circle (1) is (-g,-f) or (1/2,±√2)
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