JEE Main 2020 · 6 September · evening · Single correct · +4 / −1

The centre of the circle passing through the point (0, 1) and touching the parabola
y = x² at the point (2, 4) is :

  1. (a) ( 6/5,53/10 )
  2. (b) ( 3/10,16/5 )
  3. (c) ( (- 53)/10,16/5 )
  4. (d) ( (- 16)/5,53/10 )
Answer and solution

Answer (d) ( (- 16)/5,53/10 )

Solution

Maths sir · 5 steps

The centre must sit on the normal at (2, 4), and it must be equally far from both points.

  1. Centre on normal. A circle touching the parabola at (2, 4) shares its tangent there, so its centre lies on the normal at that point. For y = x², the slope is dy/dx = 2x = 4 at x = 2, so the normal has slope -1/4.
  2. Normal line. The normal is y - 4 = -1/4(x - 2). A centre (h, k) on it satisfies k = 9/2 - h/4.
  3. Equal radii. Both (0, 1) and (2, 4) are on the circle, so they are the same distance from (h, k). Squaring both distances and cancelling h² and k² gives a straight line.

    h² + (k-1)² = (h-2)² + (k-4)² ⇒ 4h + 6k = 19

  4. Solve together. Substitute k = 9/2 - h/4: 4h + 27 - 3h/2 = 19, so 5h/2 = -8 and h = -16/5. Then k = 9/2 + 4/5 = 53/10.
  5. So. The centre is (-16/5, 53/10), which is option (d). As a check, its squared distance to each point is 28.73.

The idea to keep. When a circle touches a curve at a point, its centre lies on the normal at that point. Combine that line with one more condition, such as equal distances, and you have the centre.

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

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Asked about 120 times in JEE Main since 2020. This question is from 2020.