Question

Prove that in two concentric circles, the chord of the larger circle which touches the smaller circle, is bisected at the point of contact. Hence, find the length of the chord of the larger circle which touches the smaller circle. if the diameter of concentric circles are 10 cm and 8 cm respectively.

Answer 1

Part - 1

Prove that in two is bisected at the point of contact.

 
Method:- 1

Given: Let two concentric circles be C₁ and C₂ with centre O.

AB be chord of the larger circle C₁ which touches the smaller circle C₂ at point P.

To prove: chord AB is bisected at point of contact (P) i.e. AP = BP

Since AB is tangent to smaller circle C₂,

OP ⊥ AB   ...(Tangent at any point of circle is perpendicular to the radius through point of contact)

Now, AB is a chord of bigger circle C₁ and OP ⊥ AB.

As perpendicular from the centre bisects the chord.

∴ OP would be bisector of chord AB.

AP = BP

Hence, proved.

Method:- 2

Since OP ⊥ AB,

∠OPA = ∠OPB = 90°

Hence, ΔOPA and ΔOPB are right triangles.

Join OA and OB.

 
In right traingle OAP,

H² = B² + P²

(OA)² = (OP)² + (AP)²

OA² – OP² = (AP)²

(AP)² = OA² – OP²   ...(i)

In right traingle OPB,

H² = B² + P²

(OB)² = (OP)² + (PB)²

(OA)² = (OP)² + (BP)²   ...(∴ OA = OB radii)

OA² – OP² = (BP)²

(BP)² = OA² – OP²   ...(ii)

From equations (i) and (ii),

AP² = BP²

AP = BP

Hence, proved.

Part - 2

Diameter of C₂ = 8 cm

r₁ = 4 cm

Diameter of C₁ = 10 cm

r₂ = 5 cm

Length AB = (?)

In ΔOPB, H² = B² + P²

5² = B² + 4²

25 – 16 = B² = 9

B² = 9

B = 3

PB = 3

AB = 3 + 3

AB = 6 cm