Let circle be 2x(x−a)+y(2y−b)=0,a≠0,b≠0. 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 Aa2>2b2 The equation of circle is, (x−a/2)2+(y−b/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 x−axis.
Let A(t,0) be mid point of PQ. Then, 2t=3a2+rcosθ,0=3b4+rsinθ⇒rcosθ=2t−3a2,rsinθ=−3b4∴r2=(2t−3a2)2+9b216
Putting the values of r in above equation, we get t2−32at+(a22+b28)=0
This will give two distinct values of t for points A,
if 94a2−4(a22+b28)>0 ⇒a24−b22>0 ⇒a2>2b2