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Question

The equation of the circle which passing through origin and cuts the circles x2+y2−4x−6y−3=0 and x2+y2−8y+12=0 orthogonally is

A
x2+y2+6x3y=0
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B
x2+y2+7x+y=0
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C
x2+y2+x+7y=0
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D
None of these
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Solution

The correct option is A x2+y2+6x3y=0
Let the required circle be x2+y2+2gx+2fc+c=0 ...(1)
As the circle passes through (0,0), we get c=0.
This circle is orthogonal to two given circles, we apply the condition 2gg1+2ff1=c+c1 to get
2g(0)8f+c+12 and 4g6f=c3
8f=12 and 4g6f=3
f=32, g=3
From (1), we get x2+y2+6x3y=0

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