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Question

Let circle be 2x(xa)+y(2yb)=0,a0,b0. Then the condition on a and b if two chords each bisected by the x-axis, can be drawn to the circle from (a,b2), is

A
a2>2b2
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B
a2<2b2
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C
a2=2b2
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D
None of these
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Solution

The correct option is A a2>2b2
The equation of circle is,
(xa/2)2+(yb/4)2=a24+b216
Radius, r=a24+b216
So any point Q(a2+rcosθ,b4+rsinθ)
Let chord PQ be drawn from P(a,b2) such that it is bisected by xaxis.
Let A(t,0) be mid point of PQ. Then,
2t=3a2+rcosθ, 0=3b4+rsinθrcosθ=2t3a2, rsinθ=3b4r2=(2t3a2)2+9b216

Putting the values of r in above equation, we get
t232at+(a22+b28)=0
This will give two distinct values of t for points A,
if 94a24(a22+b28)>0
a24b22>0
a2>2b2

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