Question

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

• A. 5√6cm
• B. 6√6cm
• C. 4√6cm
• D. 8√6cm

Answer 1

The correct option is C ($4\sqrt{6}$ cm)

Two circles touch internally at S, A and B be the centres of the bigger and smaller circle respectively.

The perpendicular bisector PQ bisects AB and meets the circle at P and Q.

To find PQ we have to join PA, AC

With given radii, we find

AS = 5cm BS = 3cm

AB = 5 -3 = 2 cm and AC = 1 cm

[perpendicular bisector the chord]

PA = radius of bigger circle = 5 cm

in right triangle ACP,

${PC}^2$ = ${PA}^2$ - ${AC}^2$ [By Pythagoras Theorem]

${PC}^2$ = $5^2$ - $1^2$

${PC}^2$ = 25 - 1 = 24

PC = $\sqrt{24}$

PC= $2\sqrt{6}$

PQ = 2 PC

PQ = $4\sqrt{6}$