JEE Main 2022 · 24 June · morning · Single correct · +4 / −1
Let x² + y² + Ax + By + C = 0 be a circle passing through (0, 6) and touching the parabola y = x² at (2, 4). Then A + C is equal to ___________.
- (a) 16
- (b) 88/5
- (c) 72
- (d) -8
Answer and solution
Answer (a) 16
Solution
Maths sir · 5 steps
A circle touching the parabola at a point touches its tangent there, so build the circle from that tangent.
- Tangent line. For y = x² the slope is 2x = 4 at x = 2. The tangent at (2, 4) is y - 4 = 4(x - 2), that is 4x - y - 4 = 0. The circle touches the parabola at (2, 4), so it also touches this line there.
- Family of circles. Every circle touching the line 4x - y - 4 = 0 at (2, 4) is the point circle at (2, 4) plus λ times the line.
(x-2)² + (y-4)² + λ(4x - y - 4) = 0
- Use the given point. Putting in (0, 6) gives 4 + 4 + λ(0 - 6 - 4) = 0, so 8 - 10λ = 0 and λ = 4/5.
- Find A and C. Expanding, the coefficient of x is A = -4 + 4λ = -4 + 16/5 = -4/5. The constant term is C = 4 + 16 - 4λ = 20 - 16/5 = 84/5.
- So. A + C = -4/5 + 84/5 = 80/5 = 16, which is option (a).
The idea to keep. When a circle touches a line or curve at a known point, write Sₚₒᵢₙₜ + λ L = 0 with L the tangent there. One more condition fixes λ.
Answer from the NTA final answer key. Solution by PrepWiser.