Question

Two concentric circles are of radii $5cm$ and $3cm$.

Find the length of the chord of the larger circle (in $cm$) which touches the smaller circle.

Answer 1

Step 1: Finding the length of $PS$:

$PQ$ is a chord of a larger circle and a tangent of a smaller circle.

Tangent $PQ$ is perpendicular to the radius at the point of contact $S$.

Therefore, $\angle \mathrm{OSP}=90°$

In $\Delta OSP$ (Right-angled triangle)

By the Pythagoras Theorem,

$\begin{array}{rcl}O{P}^2& =& O{S}^2+S{P}^2\\ 5^2& =& 3^2+S{P}^2\\ S{P}^2& =& 25–9\\ S{P}^2& =& 16\\ SP& =& \pm 4\end{array}$

$SP$ is the length of the tangent and cannot be negative

Hence, $\mathrm{SP}=4\mathrm{cm}.$

Step 2: Finding the length of $QS$:

$QS=SP$ (Perpendicular drawn from centre bisects the chord considering $\mathrm{QP}$ to be the larger circle’s chord)

Therefore, $\mathrm{QS}=\mathrm{SP}=4\mathrm{cm}$

Step 3: Finding the length of $PQ$

From figure, we have

Length of the chord$PQ=QS+SP=4+4$

$PQ=8cm$

Therefore, the length of the chord of the larger circle is $8cm.$