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Question

The equation of the circle which orthogonally intersects the circles x2+y22x+3y7=0, x2+y2+5x5y+9=0 and x2+y2+7x9y+29=0, is

A
x2+y216x8y12=0
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B
x2+y216x18y4=0
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C
x2+y2+16x18y+4=0
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D
x2+y2+16x+18y+12=0
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Solution

The correct option is B x2+y216x18y4=0
S1:x2+y22x+3y7=0S2:x2+y2+5x5y+9=0S3:x2+y2+7x9y+29=0

Radical axis of S1 and S2 is
S2S1=07x8y+16=0(1)

Radical axis of S2 and S3 is
S3S2=0x2y+10=0(2)
Radical center is point of intersection of equation (1) and (2), so
C=(8,9)

Hence, the equation of the required circle is
(x8)2+(y9)2=((S1))2(x8)2+(y9)2=(64+8116+277)x2+y216x18y4=0


Alternate Solution:
Let the required circle be
x2+y2+2gx+2fy+c=0 (1)

Since, it is orthogonal to three given circles respectively, therefore
2g×(1)+2f×32=c72g+3f=c7(2)2g×52+2f×(52)=c+95g5f=c+9(3)2g×72+2f×(92)=c+297g9f=c+29(4)
Solving equations (2),(3) and (4) we get
g=8, f=9, c=4
Substituting the values of g, f, c in (1) then required circle is
x2+y216x18y4=0

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