5 Given a circle $(x-14)^{2}+(y-12)^{2}=36^{2}$, a point $C(4,2)$ inside the circle, and two moving points $A, B$ on the circumference, such that $\angle A C B=90^{\circ}$. The equation of the locus of the midpoint of the hypotenuse $A B$ is $\qquad$ .
$5 \quad(x-9)^{2}+(y-7)^{2}=13 \times 46$.
The center of the circle is $E(14,12)$. Complete the rectangle $A C B D$. Let the midpoint of $A B$ be $M$, then by the median formula
$$
E D^{2}+E C^{2}=E A^{2}+E B^{2}=2 R^{2} .
$$

Let the midpoint of $C E$ be $F$, then the coordinates of $F$ are $(9,7)$, and
$$
F M=\frac{1}{2} D E=\frac{1}{2} \sqrt{2 R^{2}-E C^{2}}=\frac{1}{2} \sqrt{2 \times 36^{2}-10^{2}-10^{2}}=\sqrt{13 \times 46} .
$$

Therefore, the locus of $M$ is a circle with center $F(9,7)$ and radius $\sqrt{13 \times 46}$, the equation of which is
$$
(x-9)^{2}+(y-7)^{2}=13 \times 46 .
$$