Question

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

• A. $a^2>2b^2$
• B. $a^2<2b^2$
• C. $a^2=2b^2$
• D. None of these

Answer 1

The correct option is A $a^2>2b^2$

The equation of circle is,
$(x-a/2{)}^2+(y-b/4{)}^2=\frac{a^2}{4}+\frac{b^2}{16}$
Radius, $r=\sqrt{\frac{a^2}{4}+\frac{b^2}{16}}$
So any point $Q\equiv (\frac{a}{2}+r\cos \theta ,\frac{b}{4}+r\sin \theta )$
Let chord $PQ$ be drawn from $P(a,\frac{b}{2})$ such that it is bisected by $x-$axis.
Let $A(t,0)$ be mid point of $PQ.$ Then,
$2t=\frac{3a}{2}+r\cos \theta ,0=\frac{3b}{4}+r\sin \theta \Rightarrow r\cos \theta =2t-\frac{3a}{2},r\sin \theta =-\frac{3b}{4}\therefore r^2={(2t-\frac{3a}{2})}^2+\frac{9b^2}{16}$

Putting the values of $r$ in above equation, we get
$t^2-\frac{3}{2}at+(\frac{a^2}{2}+\frac{b^2}{8})=0$
This will give two distinct values of $t$ for points $A,$
if $\frac{9}{4}{a}^2-4(\frac{a^2}{2}+\frac{b^2}{8})>0$
$\Rightarrow \frac{a^2}{4}-\frac{b^2}{2}>0$
$\Rightarrow a^2>2b^2$