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Question

The equation of the image of the circle x2+y26x4y=0 in the bisector of 2nd and 4th quadrant is

A
x2+y24x6y=0
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B
x2+y2+4x6y=0
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C
x2+y24x+6y=0
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D
x2+y2+4x+6y=0
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Solution

The correct option is D x2+y2+4x+6y=0
Given circle is x2+y26x4y=0
Centre is C=(3,2)
Radius is r=9+40=13
The bisector of 2nd and 4th quadrant is
x+y=0

Let new center will be C(h,k), then
Now, the image if the centre is
h31=k21=2(3+2)2(h,k)=(2,3)

Hence, the required equation of the circle is
(x+2)2+(y+3)2=13x2+y2+4x+6y=0

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