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JEE Main 2018

JEE Main 2018 Mathematics question on Circles

If the tangent at (1, 7) to the curve x2=y−6x^2 = y - 6 touches the circle x2+y2+16x+12y+c=0x^2 + y^2 + 16x + 12y + c = 0, then the value of cc is
  1. A.
    195
  2. B.
    185
  3. C.
    85
  4. D.
    95
Show answer and solution

Correct option: D

Solution

The tangent to x² = y - 6 at (1, 7): differentiating, 2x = y', so slope = 2. Tangent line: y - 7 = 2(x - 1) → 2x - y + 5 = 0.

The circle x² + y² + 16x + 12y + c = 0 has center (-8, -6) and radius r = √(64 + 36 - c) = √(100 - c).

For the line to touch the circle, distance from center to line = radius:
|2(-8) - (-6) + 5| / √(4+1) = √(100 - c)
|-16 + 6 + 5| / √5 = √(100 - c)
|-5| / √5 = √(100 - c)
5/√5 = √5 = √(100 - c)

So 100 - c = 5, giving c = 95.

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