Question

If two chords of the circle $x^2+y^2-ax-by=0$, drawn from the point $(a,b)$ is divided by the x-axis in the ratio $2:1$, then:

• A. $a^2>3b^2$
• B. $a^2<3b^2$
• C. $a^2>4b^2$
• D. $a^2<4b^2$

Answer 1

The correct option is A $a^2>3b^2$

Let $(p,q)$ be the other end of the chord.

Then a point that divides the chord in the ratio $2:1$ is given by $(\frac{2p+a}{3},\frac{2q+b}{3})$.

If this lies on the $x-$axis then $2q+b=0$ or $q=-\frac{b}{2}$

Since $(p,q)$ lies on the given circle, we can write $p^2+{(-\frac{b}{2})}^2-ap-b(-\frac{b}{2})=0$ which is quadratic in $p$.

That is, $p^2-ap+\frac{3}{4}{b}^2=0$

$\Rightarrow p=\frac{1}{2}(a\pm \sqrt{a^2-3b^2})$ from which for real $p$ with two distinct solutions we must have $a^2>3b^2$