Eight, the moving circle is tangent to the circle $x^{2}+y^{2}=1$, and also tangent to the $y$-axis. Find the equation of the locus of the center of the moving circle, and draw the graph.



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Eight, Solution Let the coordinates of the center $P$ of the moving circle be $(x, y)$.

If the moving circle is externally tangent to the circle $x^{2}+y^{2}=1$, then
$$
\sqrt{x^{2}+y^{2}}-1=|x|
$$

If the moving circle is internally tangent to the circle $x^{2}+y^{2}=1$, then
$$
1-\sqrt{x^{2}+y^{2}}=|x|
$$
From (1), we get $\sqrt{x^{2}+y^{2}}=|x|+1$.
Squaring both sides, we get $x^{2}+y^{2}=x^{2}+2|x|+1$.
Thus, $\mathrm{y}^{2}=2|\mathbf{x}|+1$,
which means $\mathrm{y}^{2}= \pm 2 \mathrm{x}+1$.
From (2), we also get $\mathrm{y}^{2}= \pm 2 \mathrm{x}+1$.
In summary, the equations of the trajectory of the center of the moving circle are
$$
y^{2}=2 x+1 \text { and } y^{2}=-2 x+1
$$