10. As shown in Figure 3, point $A$ is on the semicircle $x^{2}+y^{2}=1(y \geqslant 0)$, and points $B(-3,-1), C(2,-1)$. Find the equation of the locus of the orthocenter $H$ of $\triangle A B C$, and sketch its approximate graph.
10. Let point $H(x, y)$.

From $A H \perp B C$, point $A$ is on the upper semicircle, we know $A\left(x, \sqrt{1-x^{2}}\right)(-1 \leqslant x \leqslant 1)$.
From $B H \perp C A$
$$
\Rightarrow \frac{y+1}{x+3} \cdot \frac{\sqrt{1-x^{2}}+1}{x-2}=-1 \text {. }
$$

This leads to the trajectory equation of point $H$ as
$$
y=\frac{6-x-x^{2}}{1+\sqrt{1-x^{2}}}-1(-1 \leqslant x \leqslant 1) \text {. }
$$

Using a TI graphing calculator, the approximate image of the trajectory is shown in Figure 5.