AMC 12 B, 2024

Answer (C): Adding the two equations and then completing the squares gives
$$2(x - 4)^2 + 2(y - 1)^2 = h + k + 32 + 2.$$
To ensure a real solution, it follows that $h + k$ is at least $-34$. This solution can be obtained by setting $x = 4$ and $y = 1$, in which case $h = 4^2 + 1^2 - 6 \cdot 4 - 8 \cdot 1 = -15$ and $k = 4^2 + 1^2 - 10 \cdot 4 + 4 \cdot 1 = -19$. The requested minimum is therefore $-34$.

Completing the squares gives
$$(x - 3)^2 + (y - 4)^2 = h + 25$$
and
$$(x - 5)^2 + (y + 2)^2 = k + 29.$$
Thus the graphs of these two equations are circles with centers at $(3, 4)$ and $(5, -2)$. The values of $h$ and $k$ are minimized when the two circles are externally tangent and have equal radii, that is, when the radii are half the distance between the two centers of the circles. See the note for further justification.

Thus the radii are both $\frac{1}{2} \cdot \sqrt{(3-5)^2 + (4+2)^2} = \sqrt{10}$. Therefore $h + 25 = k + 29 = 10$, so $h + k = 20 - 25 - 29 = -34$. The (unique) solution of the system is $(x, y, h, k) = (4, 1, -15, -19)$.