Question

Two concentric circles of radii a and b (a > b) are given. Find the length of the chord of the larger circle which touches the smaller circle.

Answer 1

It is given that AC is the tangent to the smaller circle. We know that the tangent to a circle is perpendicular to the radius through the point of contact.

∴ OB ⊥ AC

In ∆OBA,

\[\left( OB \right)^2 + \left( AB \right)^2 = \left( OA \right)^2 \left[ \text{Using Pythagoras theorem} \right]\]
\[ \Rightarrow b^2 + \left( AB \right)^2 = a^2 \]
\[ \Rightarrow \left( AB \right)^2 = a^2 - b^2 \]
\[ \Rightarrow AB = \sqrt{a^2 - b^2}\]

Since AC is the chord of the larger circle,

∴ AB = BC    (Perpendicular from the centre of a circle to the chord, bisects the chord.)

∴ AC = 2AB = 2

\[\sqrt{a^2 - b^2}\]

Thus, the length of the chord is 2

\[\sqrt{a^2 - b^2}\]