Question

The length of the common chord of the circles $x^2+y^2+px=0$ and $x^2+y^2+qy=0$ is

• A. $\frac{2pq}{\sqrt{p^2+q^2}}$
• B. $\frac{pq}{2\sqrt{p^2+q^2}}$
• C. $\frac{pq}{\sqrt{p^2+q^2}}$
• D. $\frac{pq}{p^2+q^2}$

Answer 1

Answer: (C)

$\frac{pq}{\sqrt{p^2+q^2}}$

$S_1:x^2+y^2+px=0\Rightarrow C_1\equiv (-\frac{p}{2},0)$ & $r_1=\frac{p}{2}$
$S_2:x^2+y^2+qy=0\Rightarrow C_2\equiv (0,-\frac{q}{2})$ & $r_2=\frac{q}{2}$
As the circles have common chord, so $S_1$ and $S_2$ intersects each other.
Equation of common chord, $AB$ is $S_2-S_1=0$

$\Rightarrow px=qy$
So,

Distance of midpoint of $AB$, i.e $P$ from $C_1$ is
$P{C}_1=\left|\frac{\frac{-p^2}{2}}{\sqrt{p^2+q^2}}\right|=\left|\frac{p^2}{2\sqrt{p^2+q^2}}\right|$
So, using Pythagorus Theorem in $\Delta AP{C}_1$, we get

$AP=\sqrt{{A{C}_1}^2-{P{C}_1}^2}$
$AP=\sqrt{\frac{p^2}{4}-\frac{p^4}{4(p^2+q^2)}}$

$\because A{C}_1=r_1$
$AP=\frac{p}{2}\frac{q}{\sqrt{p^2+q^2}}$
So, length of common chord, $AB$ is
$AB=2AP=\frac{pq}{\sqrt{p^2+q^2}}$