Question

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

Answer 1

So, in figure we can see that AC is radius of bigger circle and BC is radius of circle.

As in given question

AC = 5 cm and BC = 3 cm

AC - BC = AB

5 cm - 3 cm = AB

2 cm = AB

Now, PQ is perpendicular bisector of AB so AD = 1 cm.

Now, since ∆ADP is right triangle at D so,

AP² = PD² + AD²

5² - 1² = PD²

√24 = PD

2√6 cm = PD

Now, PQ is chord which intersect AC that passes through centre so AC is perpendicular bisector on PQ.

PD = QD

PD + QD = PQ

2PD = PQ

2(2√6) = PQ

4√6 = PQ.

Hence length of PQ is 4√6.

Answer 2

$\Large{\underline{\underline{\bf{Given:-}}}}$

Two circles with centres A and B and of radii 5 cm and 3 cm touch each other internally.

$\Large{\underline{\underline{\bf{To \: Find:-}}}}$

Length of PQ.

$\Large{\underline{\underline{\bf{Solution:-}}}}$

If two circles touch internally then distance between their centers is equal to the difference of their radii.

so, AB = (5 - 3) cm

AB= 2 cm

Also , the common chord PQ is the perpendicular bisector of AB.

∴ AC = CB = 1 cm

From right triangle ACP , we have

⟹ AP² = AC² + CP²

⟹ 5² = 1² + CP²

⟹ CP² = 25 - 1

⟹ CP = √24cm

Hence , PQ = 2CP

⟹PQ = 2√24 cm

⟹PQ = 4√6cm

$\Large{\underline{\underline{\bf{Thanks}}}}$