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Question

Consider three circles whose equations are x2+y2+3x+2y+1=0, x2+y2x+6y+5=0 and x2+y2+5x8y+15=0, then

A
equation of the circle which is orthogonal to given circles isx2+y26x4y14=0
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B
equation of the circle which is orthogonal to given circles isx2+y26x+4y13=0
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C
radical centre of the circles is (3,2)
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D
radical centre of the circles is(3,2)
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Solution

The correct option is C radical centre of the circles is (3,2)
Given circles are
S1:x2+y2+3x+2y+1=0
S2:x2+y2x+6y+5=0
S3:x2+y2+5x8y+15=0

Equations of two radical axes are
S1S2=04x4y4=0xy1=0

And
S2S3=06x+14y10=03x7y+5=0

Solving them, we get the radical centre as (3,2)
Also, if r is the length of the tangent drawn from the radical centre (3,2) to any one of the given circles, say S1, we have
r=S1=32+22+3.3+2.2+1r=27

Hence, (3,2) is the centre and 27 is the radius of the circle intersecting them orthogonally.
Its equation is
(x3)2+(y2)2=27x2+y26x4y14=0

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