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Question

If the abscissa and ordinates of two points P and Q are the roots of the equations x2+2ax−b2=0 and x2+2px−q2=0 respectively, then the equation of the circle with PQ as diameter is

A
x2+y2+2ay+2pxb2q2=0
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B
x2+y2+2ax+2pyb2q2=0
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C
x2+y2+2(a+p)xb2q2=0
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D
x2+y2+2(a+p)yb2q2=0
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Solution

The correct option is B x2+y2+2ax+2pyb2q2=0
Let P(x1,y1),Q(x2,y2) be end points of diameter.
Given
x1,x2 are the roots of x2+2axb2=0
y1,y2 are the roots of x2+2pxq2=0
x1+x2=2a, x1x2=b2
y1+y2=2p, y1y2=q2
Equation of circle having PQ as diameter is
(xx1)(xx2)+(yy1)(yy2)=0
x2(x1+x2)x+x1x2+y2(y1+y2)y+y1y2=0
x2+y2+2ax+2pyb2q2=0

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