# 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.

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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.

by (65.0k points)
selected 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.

by (13.5k points)

Solution: 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.

by (65.0k points)
+2
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?