Answer :

2r

Given - ∠ BPA = 60⁰

To find - length of OP

Properties -

1. Lengths of the two tangents drawn from an external point to a circle are equal.

2. In 30⁰-60⁰-90⁰ triangles, the side opposite to 30⁰ is half of the hypotenuse and the side opposite to 60⁰ is times the hypotenuse.

3. The radius from the centre of the circle to the point of tangency is perpendicular to the tangent line.

Answer -

As we know that lengths of tangents drawn from the same external point to the circle are same.

∴ PA = PB ………(1)

The angle between the two tangents is 60⁰.

∴ ∠ BPA = 60⁰ ………(2)

Therefore, for ∆POB & ∆POA,

PB = PA ………from (1)

OB = OA = r ………..radii of same circle

And OP is the common side.

………by SSS test of similarity

∴ ∠ BPO = ∠ APO ……….corresponding angles of similar triangles ……….(3)

But ∠ BPA = ∠ BPO + ∠ APO

∴ 60⁰ = ∠ BPO + ∠ BPO ……….from (2) & (3)

∴ 60⁰ = 2 .∠ BPO

∴ ∠ BPO = 30⁰ = ∠ APO

As tangents are always perpendicular to the radius

∴ ∠ PBO = ∠ PAO = 90⁰

Therefore, ∆POB & ∆POA are 30⁰-60⁰-90⁰.

Therefore, by the property of 30⁰-60⁰-90⁰ triangle, the side opposite to 30⁰ is half of hypotenuse.

In ∆POB, ∠ BPO = 30⁰ and side opposite to ∠ BPO is OB, therefore,

∴ OP = 2r

Hence, the length of OP is 2r units.

**OR**

6 cm

In the figure below, EF is the chord for outer circle, and it is tangent for inner circle at point D.

Given – radius of inner circle = AD = 4 cm

Radius of outer circle = AE = AF = 5 cm

To Find – length of chord EF

Properties –

1. The radius from the centre of the circle to the point of tangency is perpendicular to the tangent line.

Answer –

As radius from the centre of the circle to the point of tangency is perpendicular to the tangent line.

∴ ∠ ADE = ∠ ADF = 90⁰ ………(1)

AE = AF = 5 cm ………(2) (given)

AD = 4 cm ………(3) (given)

In right angled triangle ADE,

By Pythagoras Theorem,

AE^{2} = AD^{2} + ED^{2}

∴ 5^{2} = 4^{2} + ED^{2}

∴ 25 = 16 + ED^{2}

∴ ED^{2} = 9

∴ ED = 3 ………(4)

Similarly, in right angled triangle ADF,

AF^{2} = AD^{2} + FD^{2}

∴ 5^{2} = 4^{2} + FD^{2}

∴ 25 = 16 + FD^{2}

∴ FD^{2} = 9

∴ FD = 3 ………(5)

Now, EF = ED + FD

∴ EF = 3 + 3 = 6 cm

Therefore, length of chord is 6 cm.

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