Question:
Let \(x^2 + y^2 + Ax + By + C = 0\) be a circle passing through (0, 6) and touching the parabola \(y = x^2\) at (2, 4). Then A + C is equal to __________.
Let \(x^2 + y^2 + Ax + By + C = 0\) be a circle passing through (0, 6) and touching the parabola \(y = x^2\) at (2, 4). Then A + C is equal to __________.
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Write the circle as the family \(S + \lambda L = 0\), where S is the point circle at (2, 4) and L the tangent to the parabola there.
Updated On: Nov 4, 2026
- 16
- 88/5
- 72
- \(-8\)
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The Correct Option is A
Solution and Explanation
Tangent to \(y = x^2\) at (2, 4): slope \(= 2x = 4\), line \(L: 4x - y - 4 = 0\).
Circle touching L at (2, 4): \((x-2)^2 + (y-4)^2 + \lambda(4x - y - 4) = 0\).
Through (0, 6): \(4 + 4 + \lambda(-6 - 4) = 0 \Rightarrow \lambda = \frac45\).
Expand: \(x^2 + y^2 + \left(-4 + \frac{16}{5}\right)x + \left(-8 - \frac45\right)y + \left(20 - \frac{16}{5}\right) = 0\), so \(A = -\frac45\), \(B = -\frac{44}{5}\), \(C = \frac{84}{5}\).
\(A + C = \frac{80}{5} = 16\).
Answer: (1) 16
Circle touching L at (2, 4): \((x-2)^2 + (y-4)^2 + \lambda(4x - y - 4) = 0\).
Through (0, 6): \(4 + 4 + \lambda(-6 - 4) = 0 \Rightarrow \lambda = \frac45\).
Expand: \(x^2 + y^2 + \left(-4 + \frac{16}{5}\right)x + \left(-8 - \frac45\right)y + \left(20 - \frac{16}{5}\right) = 0\), so \(A = -\frac45\), \(B = -\frac{44}{5}\), \(C = \frac{84}{5}\).
\(A + C = \frac{80}{5} = 16\).
Answer: (1) 16
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