Question

In the adjoining figure, two circles with centres at A and B, and of radii 5 cm and 3 cm touch each other internally. If the perpendicular bisector of AB meets the bigger circle in P and Q, find the length of PQ.

Answer 1

Two circles with centres A and B of respective radii 5 cm and 3 cm touch each other internally.
The perpendicular bisector of AB meets the bigger circle at P and Q.
Join AP.

Let PQ intersect AB at point L.
Here, AP = 5 cm
Then AB = (5 - 3) cm = 2 cm
Since PQ is the perpendicular bisector of AB, we have:
$AL\mathit{=}\left(\frac{\mathit{A}\mathit{B}}{\mathit{2}}\right)=\left(\frac{2}{2}\right)=1\mathrm{cm}$
Now, in right angled ΔPLA, we have:
AP² = AL² + PL²
⇒ PL² = AP² - AL²
= 5² - 1²
= 25 - 1 = 24
⇒ $\mathrm{PL}=\sqrt{24}=2\sqrt{6}\mathrm{cm}$
Thus PQ = 2 × PL
= $\left(2\times 2\sqrt{6}\right)=4\sqrt{6}\mathrm{cm}$
Hence, the required length of PQ is $4\sqrt{6}\mathrm{cm}$.