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 ______

The circle $$x^2 + y^2 + Ax + By + C = 0$$ passes through $$(0, 6)$$ and touches the parabola $$y = x^2$$ at $$(2, 4)$$.

Since the circle passes through $$(0, 6)$$:

$$ 0 + 36 + 0 + 6B + C = 0 $$

$$ 6B + C = -36 \quad \cdots (1) $$

Since the circle passes through $$(2, 4)$$:

$$ 4 + 16 + 2A + 4B + C = 0 $$

$$ 2A + 4B + C = -20 \quad \cdots (2) $$

The circle touches the parabola $$y = x^2$$ at $$(2, 4)$$, so they have the same tangent at that point.

The slope of the parabola at $$(2, 4)$$: $$\frac{dy}{dx} = 2x = 4$$.

For the circle, differentiating $$x^2 + y^2 + Ax + By + C = 0$$ implicitly:

$$ 2x + 2y\frac{dy}{dx} + A + B\frac{dy}{dx} = 0 $$

At $$(2, 4)$$ with $$\frac{dy}{dx} = 4$$:

$$ 4 + 8(4) + A + 4B = 0 $$

$$ 4 + 32 + A + 4B = 0 $$

$$ A + 4B = -36 \quad \cdots (3) $$

Solve the system of equations.

From (3): $$A = -36 - 4B$$

Substitute into (2):

$$ 2(-36 - 4B) + 4B + C = -20 $$

$$ -72 - 8B + 4B + C = -20 $$

$$ -4B + C = 52 \quad \cdots (4) $$

From (1): $$C = -36 - 6B$$. Substitute into (4):

$$ -4B + (-36 - 6B) = 52 $$

$$ -10B - 36 = 52 $$

$$ -10B = 88 $$

$$ B = -\frac{88}{10} = -\frac{44}{5} $$

$$ C = -36 - 6\left(-\frac{44}{5}\right) = -36 + \frac{264}{5} = \frac{-180 + 264}{5} = \frac{84}{5} $$

$$ A = -36 - 4\left(-\frac{44}{5}\right) = -36 + \frac{176}{5} = \frac{-180 + 176}{5} = -\frac{4}{5} $$

Compute $$A + C$$:

$$ A + C = -\frac{4}{5} + \frac{84}{5} = \frac{80}{5} = 16 $$

The answer is Option A: 16.