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Question

The equation of the image of the circle x2+y2−6x−4y=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=0Given circle is x2+y2−6x−4y=0 Centre is C=(3,2) Radius is r=√9+4−0=√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 h−31=k−21=−2(3+2)2⇒(h,k)=(−2,−3) Hence, the required equation of the circle is (x+2)2+(y+3)2=13∴x2+y2+4x+6y=0

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