The area (in sq units) of the smaller of the two circles that touch the parabola, at the point and the -axis is
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: A
Solution
The tangent to y² = 4x at (1,2) is y·2 = 2(x+1), i.e., x - y + 1 = 0. Any circle tangent to the parabola at (1,2) has its center on the normal at (1,2). The normal slope is -1 (negative reciprocal of tangent slope 1), so the normal line is y - 2 = -1(x - 1), giving center at (1+t, 2-t) for some t, with radius r = t√2 (distance from center to (1,2)).
The circle also touches the X-axis, so the distance from center to X-axis equals the radius:
|2 - t| = t√2
Case 1: 2 - t = t√2 → t = 2/(1+√2) = 2(√2-1)
Case 2: t - 2 = t√2 → t = 2/(1-√2) = -2(√2+1) (negative, gives larger circle with t positive meaning center below)
For the smaller circle, take t = 2(√2-1):
r = t√2 = 2√2(√2-1) = 4 - 2√2
Area = πr² = π(4-2√2)² = π(16 - 16√2 + 8) = π(24 - 16√2) = 8π(3 - 2√2)
The circle also touches the X-axis, so the distance from center to X-axis equals the radius:
|2 - t| = t√2
Case 1: 2 - t = t√2 → t = 2/(1+√2) = 2(√2-1)
Case 2: t - 2 = t√2 → t = 2/(1-√2) = -2(√2+1) (negative, gives larger circle with t positive meaning center below)
For the smaller circle, take t = 2(√2-1):
r = t√2 = 2√2(√2-1) = 4 - 2√2
Area = πr² = π(4-2√2)² = π(16 - 16√2 + 8) = π(24 - 16√2) = 8π(3 - 2√2)
[JEE Main 2019 · 9 April · Shift II]