Question

The length of the common chord of the two circles $(x-a)^{2}+y^{2}=a^{2}$ and
$x^{2}+(y-b)^{2}=b^{2}$, is

• A. $\frac{a\, b}{\sqrt{a^{2}+b^{2}}}$
• B. $\frac{2 \,a b}{\sqrt{a^{2}+b^{2}}}$
• C. $\frac{a+b}{\sqrt{a^{2}+b^{2}}}$
• D. $\sqrt{a^{2}+b^{2}}$

Answer 1

Answer: (B)

The equation of the common chord $P Q$ of the circles

$S_{1}: x^{2}+y^{2}-2\, a x=0$
$S_{2}: x^{2}+y^{2}-2\,b y=0$
is $S_{1}-S_{2}=0$
$\Rightarrow 2 \,b y-2\,a x=0$
$\Rightarrow a x-b y=0$
The centre of $S_{1}$ is $(a, 0)$ and radius is $a$.
The length of the perpendicular from $(a, 0)$ to $a x-b y=0$, is
$ C_{1} M=\left|\frac{a^{2}}{\sqrt{a^{2}+b^{2}}}\right|$
Now, $ P Q =2 P M=2 \sqrt{C_{1} P^{2}-C_{1} M^{2}} $
$=2 \sqrt{a^{2}-\frac{a^{4}}{a^{2}+b^{2}}}=\frac{2 a b}{\sqrt{a^{2}+b^{2}}}$