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. $\frac{ab}{\sqrt{a^2+b^2}}$
• B. $\frac{2ab}{\sqrt{a^2+b^2}}$
• 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 $PQ$ of the circles

$S_1:x^2+y^2−2ax=0$
$S_2:x^2+y^2−2by=0$
is $S_1−S_2=0$
$⇒2by−2ax=0$
$⇒ax−by=0$
The centre of $S_1$ is $(a,0)$ and radius is $a$.
The length of the perpendicular from $(a,0)$ to $ax−by=0$, is
$C_{1}M=â\frac{a^2}{\sqrt{a^2+b^2}}â$
Now, $PQ=2PM=2\sqrt{C_1{P}^2−C_1{M}^2}$
$=2\sqrt{a^2−\frac{a^4}{a^2+b^2}}=\frac{2ab}{\sqrt{a^2+b^2}}$