Question

Two circles with centers A and B, and radii 5 cm and 3 cm, touch each other internally, If the perpendicular bisector of the segment AB meets the bigger circle in P and Q; find the length of PQ.

• A. 8\sqrt{6}\) cm.
• B. 4\sqrt{6}\) cm.
• C. 2\sqrt{6}\) cm.
• D. 9\sqrt{6}\) cm.

Answer 1

The correct option is B

4\sqrt{6}\) cm.

Two circles with centres A and B touch each other at C internally.

PQ is perpendicular bisector of AB meeting the bigger circles at P and Q. Join AP.

Radius AC = 5 cm. and radius BC = 3 cm.

AB=AC - BC = 5-3=2 cm.

AP= 5 cm, AM = $\frac{1}{2}AB=\frac{1}{2}\times 2=1$cm.

Now in right $\Delta$ APM

$A{P}^2=A{M}^2+M{P}^2\Rightarrow (5{)}^2=(1{)}^2+M{P}^2$

$\Rightarrow 25=1+M{P}^2$

$\therefore M{P}^2=25-1=24=4\times 6$

$\therefore MP=\sqrt{4\times 6}=2\sqrt{6}$ cm.

$\therefore PQ=2MP=2\times 2\sqrt{6}=4\sqrt{6}$ cm.