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Question

If two distinct chords drawn from the point (p, q) on the circle x2+y2pxqy=0 (where pq0) are bisected by the x-axis, then

A
p2=q2
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B
p2=8q2
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C
p2<8q2
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D
p2>8q2
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Solution

The correct option is D p2>8q2
Let B(h, 0) be the midpoint of the chord drawn from point A(p, q).
Also, the center of the circle is C(p2,q2).
Then, we have BCAB. Therefore,
(q2)0(p2)h(q0ph)=1 ...(Product of slopes of two perpendicular lines, m1.m2=1)(qp2h)(q0ph)=12h23ph+p2+q2=0
Since two such chords exist, the above equation must have two distinct real roots, i.e.,
Discriminant > 0
9p28(p2+q2)>0or p2>8q2

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