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Question

# The equation of the circle which orthogonally intersects the circles x2+y2−2x+3y−7=0, x2+y2+5x−5y+9=0 and x2+y2+7x−9y+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+y2−16x−18y−4=0S1:x2+y2−2x+3y−7=0S2:x2+y2+5x−5y+9=0S3:x2+y2+7x−9y+29=0 Radical axis of S1 and S2 is S2−S1=0⇒7x−8y+16=0⋯(1) Radical axis of S2 and S3 is S3−S2=0⇒x−2y+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 (x−8)2+(y−9)2=(√(S1))2⇒(x−8)2+(y−9)2=(64+81−16+27−7)⇒x2+y2−16x−18y−4=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=c−7⇒−2g+3f=c−7……(2)2g×52+2f×(−52)=c+9⇒5g−5f=c+9……(3)2g×72+2f×(−92)=c+29⇒7g−9f=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+y2−16x−18y−4=0

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