17. Given a line segment $AB$ of fixed length 4, with its endpoints moving on two intersecting lines $x \pm 2y = 0$.
(1) Let the midpoint of line segment $AB$ be $G$. Find the equation of the trajectory $C$ of point $G$;
(2) If two tangents drawn from point $P$ to curve $C$ are perpendicular to each other, prove that the moving point $P$ lies on a fixed circle.
(1) Let $G(a, b)$, line $A B:\left\{\begin{array}{l}x=a+t \cos \alpha, \\ y=b+t \sin \alpha,\end{array}\right.$ substitute into the equation $x^{2}-4 y^{2}=0$, we get $\left(\cos ^{2} \alpha-4 \sin ^{2} \alpha\right) t^{2}+(2 a \cos \alpha-8 b \sin \alpha) t+a^{2}-4 b^{2}=0$.

Substitute $t_{1} t_{2}=-4 \Rightarrow \frac{a^{2}-4 b^{2}}{\cos ^{2} \alpha-4 \sin ^{2} \alpha}=-4 \Rightarrow a^{2}-4 b^{2}=\frac{16\left(a^{2}-4 b^{2}\right)}{a^{2}+16 b^{2}}$ $\Rightarrow a^{2}+16 b^{2}=16 \Rightarrow \frac{a^{2}}{16}+b^{2}=1$. Therefore, the equation of $C$ is $\frac{x^{2}}{16}+y^{2}=1$.
(2) Let $P(m, n)$, the tangent line equation is $y-n=k(x-m)$, substitute into the ellipse equation $x^{2}+16 y^{2}=16$

we get $\left(1+16 k^{2}\right) x^{2}-32 k(k m-n) x+16(k m-n)^{2}-16=0$.
$\Delta=32^{2} k^{2}(k m-n)^{2}-4 \cdot 16\left(1+16 k^{2}\right)\left[(k m-n)^{2}-1\right]=0$
$\Rightarrow\left(m^{2}-16\right) k^{2}-2 m n k+n^{2}-1=0$, noting that $k_{1}, k_{2}$ satisfy the above equation,
thus $k_{1} k_{2}=-1 \Rightarrow m^{2}+n^{2}-17=0$. Therefore, the moving point $P$ is on the fixed circle $x^{2}+y^{2}=17$.