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

$2 \sqrt { a^2 - b^2 }$

Step by Step Explanation:
1. Let $O$ be the common center of the two circles and $AB$ be the chord of the larger circle which touches the smaller circle at $C$.

As the radius of the larger circle is $a$ and that of the smaller circle is $b$, we have $OA = a$ and $OC = b$.

The situation given can be represented by the image below.
2. $AB$ is the tangent to the circle with radius $b$.
The tangent at any point of a circle is perpendicular to the radius through the point of contact.
So, $OC \perp AB.$

$AB$ is the chord of the circle with radius $a$.
The perpendicular drawn from the center of the circle to a chord bisects the chord.
$\therefore OC$ bisects $AB \implies C$ is the mid-point of $AB$.
3. Using Pythagoras' theorem in the right- angled triangle $ACO$, we have \begin{aligned} & OA^2 = OC^2 + AC^2 && \\ \implies & AC = \sqrt { OA^2 - OC^2 } = \sqrt { a^2 - b^2 } \end{aligned} $C$ is the mid-point of $AB.$ So, $AB = 2AC = 2 \sqrt { a^2 - b^2 }.$
4. Thus, the length of the chord of the larger circle which touches the smaller circle is $2 \sqrt { a^2 - b^2 }$.