Question

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

Answer 1

The given circle is $S=x^2+y^2-ax-\frac{b}{@}y=0$
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 $T=S_2$ is
$x.h+y.0-\frac{a}{2}(x+h)-\frac{b}{4}(y+0)=h^2-ah$.
It passes through $(a,b/2)$
$\therefore ah-\frac{a}{2}(a+h)-\frac{b}{4}\left(\frac{b}{2}\right)=h^2-ah$.
or $h^2-\frac{3}{2}ah+(\frac{a^2}{2}+\frac{b^2}{8})=0$
Since the values of $h$ are real and distinct we must have $B^2-4AC>0$
$\therefore \frac{9}{4}{a}^2-4\left(\frac{a^2}{2}\frac{b^2}{8}\right)>0$ or $a^2>2b^2$.