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Question

If two chords of the circle x2+y2−ax−by=0, drawn from the point (a,b) is divided by the x-axis in the ratio 2:1, then:

A
a2>3b2
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B
a2<3b2
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C
a2>4b2
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D
a2<4b2
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Solution

The correct option is A a2>3b2
Let (p,q) be the other end of the chord.

Then a point that divides the chord in the ratio 2:1 is given by (2p+a3,2q+b3).

If this lies on the xaxis then 2q+b=0 or q=b2

Since (p,q) lies on the given circle, we can write p2+(b2)2apb(b2)=0 which is quadratic in p.

That is, p2ap+34b2=0

p=12(a±a23b2) from which for real p with two distinct solutions we must have a2>3b2

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