Question

In the following figure, AB is diameter of a circle centered at O. BC is tangent to the circle at B. If OP bisects the chord AD and ∠AOP = 60°, then find m∠C.

Answer 1

Given: AB is a diameter of the circle with centre O. BC is tangent at B. OP bisects chord AD (so OP ⟂ AD and P is midpoint of AD). ∠AOP = 60°. 

Step-wise calculation:

1. Put O at (0, 0) and radius = 1 for convenience. 

Then A = (0, 1), B = (0, –1) and the tangent at B is the line y = –1.

2. OA is along the positive y-axis (direction 90°).

Since ∠AOP = 60°, the ray OP makes angle 90° – 60° = 30° with the positive x-axis.

So OP has slope `tan 30^circ = 1/sqrt(3)`. 

Thus OP: `y = (1/sqrt(3))x`.

3. OP ⟂ AD, so chord AD has slope `-sqrt(3)`.

The line through A with that slope is

`y - 1 = -sqrt(3)(x - 0)` 

⇒ `y = -sqrt(3)x + 1` 

This is the equation of chord AD and its extension AC.

4. Find the other intersection D of this line with the unit circle x² + y² = 1:

Substitute `y = -sqrt(3)x + 1` into x² + y² = 1 

⇒ `4x^2 - 2sqrt(3)x = 0`

⇒ x = 0 (gives A) or `x = sqrt(3)/2`. 

So `D = (sqrt(3)/2, -1/2)`.

5. Find C as the intersection of the extension of AD with the tangent y = –1:
Solve `-1 = -sqrt(3)x + 1` 

⇒ `x = 2/sqrt(3)`. 

So `C = (2/sqrt(3), -1)`.

6. Slope of line CA (same as AD) is `-sqrt(3)`, so the angle that CA makes with the positive x-axis is arctan `(-sqrt(3))` = –60° or 120°. 

The tangent line BC is horizontal (angle 0°).

The acute angle between BC and CA is |0° – (–60°)| = 60°.