Let and be tangents at the extremities of the diameter of a circle of radius . If and intersect at a point on the circumference of the circle, then equals
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: A
Solution
Set up coordinates: P = (-r, 0), R = (r, 0), with PQ tangent at P (vertical line x = -r) and RS tangent at R (vertical line x = r). Let PQ = a (so Q = (-r, a)) and RS = b (so S = (r, b)).
X lies on the circle, so X = (r cosθ, r sinθ). X is the intersection of PS and RQ.
Line PS: from P(-r, 0) to S(r, b). Parametrically: x = -r + 2rt, y = bt. At X: cosθ = -1 + 2t, sinθ = bt/r.
Line RQ: from R(r, 0) to Q(-r, a). Parametrically: x = r - 2rs, y = as. At X: cosθ = 1 - 2s, sinθ = as/r.
From sinθ: bt/r = as/r → bt = as. Also t = (1 + cosθ)/2, s = (1 - cosθ)/2.
So b(1 + cosθ) = a(1 - cosθ) ... no, let me use sin²θ + cos²θ = 1 with sinθ = bt/r and t = (1+cosθ)/2:
sinθ·r = b·(1+cosθ)/2 → 2r sinθ = b(1+cosθ)
Similarly: 2r sinθ = a(1-cosθ)
Multiply: 4r² sin²θ = ab(1-cos²
X lies on the circle, so X = (r cosθ, r sinθ). X is the intersection of PS and RQ.
Line PS: from P(-r, 0) to S(r, b). Parametrically: x = -r + 2rt, y = bt. At X: cosθ = -1 + 2t, sinθ = bt/r.
Line RQ: from R(r, 0) to Q(-r, a). Parametrically: x = r - 2rs, y = as. At X: cosθ = 1 - 2s, sinθ = as/r.
From sinθ: bt/r = as/r → bt = as. Also t = (1 + cosθ)/2, s = (1 - cosθ)/2.
So b(1 + cosθ) = a(1 - cosθ) ... no, let me use sin²θ + cos²θ = 1 with sinθ = bt/r and t = (1+cosθ)/2:
sinθ·r = b·(1+cosθ)/2 → 2r sinθ = b(1+cosθ)
Similarly: 2r sinθ = a(1-cosθ)
Multiply: 4r² sin²θ = ab(1-cos²
[2001]