Question

In fig., two concentric circles of radii $a$ and $b(a>b)$ are given. The chord $AB$ of larger circle touches the smaller circle at $C$. The length of $AB$ is:

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

Answer 1

The correct option is B $2\sqrt{a^2-b^2}$

Let $C_1,C_2$ be two circles of radius $a,b$ respectively.

Chord $AB$ tangent to $C_2$ at point $C$.

Join $O-B$

$OA=a=OB,OC=b$

$AB$ is tangent to $C_2$ and $OC$ perpendicular $AB$

$\therefore$ $\angle OCA={90}^o$

Using Pythagoras theorem,

$O{A}^2=O{C}^2+A{C}^2$

$⟹A{C}^2=O{A}^2-O{C}^2$

$⟹AC=\sqrt{a^2-b^2}$

Similarly, $BC=\sqrt{a^2-b^2}$

$\therefore AB=AC+CB=\sqrt{a^2-b^2}+\sqrt{a^2-b^2}$

Hence, $AB=2\sqrt{a^2-b^2}$.