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Question

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

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Solution

The given circle is S=x2+y2axb@y=0
Since the two chords are bisected by x-axis so let (h,0) be the mid-point where h has two real values.
Equation of the chord by T=S2 is
x.h+y.0a2(x+h)b4(y+0)=h2ah.
It passes through (a,b/2)
aha2(a+h)b4(b2)=h2ah.
or h232ah+(a22+b28)=0
Since the values of h are real and distinct we must have B24AC>0
94a24(a22b28)>0 or a2>2b2.

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