Let A(1, 4) and B(1, -5) be two points. Let P be a point on the circle
(x - 1)² + (y - 1)² = 1 such that (PA)² + (PB)² have maximum value, then the points, P, A and B lie on :

Solution

Maths sir · 4 steps

Put P on the circle with an angle; the sum becomes one sine term, and the maximum puts P on x = 1.

1. Parametrise P. The circle has centre (1, 1) and radius 1, so take P = (1 + cos θ, 1 + sin θ).
2. Each distance. PA² = cos² θ + (sin θ - 3)² = 10 - 6 sin θ. Also PB² = cos² θ + (sin θ + 6)² = 37 + 12 sin θ.

   PA² + PB² = 47 + 6 sin θ

3. Maximum. The sum is largest when sin θ = 1, so cos θ = 0. That gives P = (1, 2) and a maximum value of 53.
4. So. P(1, 2), A(1, 4) and B(1, -5) all have x = 1, so they lie on a straight line, option (a).

The idea to keep. For a point on a circle, write it as centre plus (r cos θ, r sin θ). Squared distances then reduce to a simple sine and cosine expression that is easy to maximise.

Answer from the NTA final answer key. Solution by PrepWiser.