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 the point $(a,\frac{b}{2})$.

• A. $a^2>2b^2$
• B. $2a^2>b^2$
• C. $a^2<2b^2$
• D. $2a^2<b^2$

Answer 1

Answer: (A)

$a^2>2b^2$

Let equation of circle be $C:2x^2+2y^2-2ax-by=0$
Point $A(2h-a,-\frac{b}{2})$ lies on the above circle.

$\therefore$ $2{(2h-a)}^2+2\frac{b^2}{4}-2a(2h-a)-b\left(\frac{-b}{2}\right)=0$
$\Rightarrow 2(4h^2-4ah+a^2)+\frac{b^2}{2}-4ah+2a^2+\frac{b^2}{2}=0$

$\Rightarrow 8h^2-12ah+4a^2+b^2=0$

$\Rightarrow$ $144a^2-4.8(4a^2+b^2)>0$ .....[$D>0$]
$\Rightarrow$ $9a^2-8a^2-2b^2>0$

$\Rightarrow$ $a^2>2b^2$