Question

Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.

Answer 1

Let O be the common centre of the two concentric circle.
Let PQ be a chord of the larger circle which touches the smaller circle at M.
Join OM and OP.
Since, the tangent at any point of a circle is perpendicular to the radius through the point of contact.
Therefore,
∠OMP = 90°
Now,
In ΔOMP, we have
OP² = OM² + PM²
[Using Pythagoras theorem]
⇒ (5)² = (3)² + PM²
⇒ 25 = 9 + PM²
⇒ PM² = 16
⇒ PM = 4 cm
Since, the perpendicular from the centre of a circle to a chord bisects the chord.
Therefore,
PM = MQ = 4 cm
∴ PQ = 2 PM = 2 x 4 = 8 cm
Hence, the required length = 8 cm.

Answer 2

Let the two concentric circles with centre O.

AB be the chord of the larger circle which touches the smaller circle at point P. 

∴ AB is tangent to the smaller circle to the point P.

⇒ OP ⊥ AB
By Pythagoras theorem in ΔOPA,
OA² =  AP² + OP²
⇒ 5² = AP² + 3²
⇒ AP² = 25 - 9
⇒ AP = 4
In ΔOPB,
Since OP ⊥ AB,
AP = PB (Perpendicular from the center of the circle bisects the chord)
AB = 2AP = 2 × 4 = 8 cm

∴ The length of the chord of the larger circle is 8 cm.

Comments

First learn how to talk. This is not the way to claim someone, and how can you say the answer is wrong.
Do you have any explanation?