#### Question

Two circle with centres 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.

#### Solution

If two circles touch internally, then distance between their centres is equal to the difference of their radii. So, AB = (5 − 3) cm = 2 cm.

Also, the common chord PQ is the perpendicular bisector of AB. Therefore, AC = CB =`1/2 AB`

= 1 cm

In right ΔACP, we have`AP^2 = AC^2 + CP^2`

⇒ `5^2 = 1^2 + CP^2`

⇒ `CP^2 = 25 - 1 = 24`

⇒ `cp = sqrt(24)= 2 sqrt(6) cm`

Now , PQ = 2 CP

= `2xx 2 sqrt(6) cm`

= `4 sqrt (6 ) cm`

Is there an error in this question or solution?

Solution Two Circle with Centres 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. Concept: Chord Properties - the Perpendicular to a Chord from the Center Bisects the Chord (Without Proof).