More generally, is it valid to define an angle created between a line and a curve?
Yes. It can be done with a little calculus. Calculus gives us a workable definition of a unique tangent vector at each point on a curve. So, we can use calculus to translate questions about curves into equivalent questions about lines.
The line $\overline{AH}$ crosses the circle at $A$ and $H$. It is intuitively plausible that the line $\overline{GH}$ from the point $G$ on the circle to the point $H$ becomes perpendicular to $\overline{AH}$ as $G$ approaches $H$. If there is a line through the circle perpendicular to $\overline{AH}$ then it must be the line that $\overline{GH}$ approaches as $G$ goes to $H$. It is possible to rigorously define such a line using the mathematically precise definitions of limit and derivative from calculus.
The equation for the curve of the circle in polar coordinates is
$(x(\theta), y(\theta))=(r\cos(\theta), r\sin(\theta))$ with $r$ is constant.
The equation for the line through the circle in polar coordinates is
$(x(r), y(r))=(r\cos(\theta), r\sin(\theta))$ with $\theta$ is constant.
The vector tangent to the circle at $\theta$ is $\vec{a}=lim_{h\rightarrow 0} \frac{(x(\theta+h), y(\theta+h))-(x(\theta), y(\theta))}{h}$.
This is a mathematically rigorous definition of the notion of the tangent to the curve at a point.
$\vec{a}=\frac{d(x(r,\theta),y(r\theta))}{d\theta}=\frac{d(r\cos(\theta),(r\sin(\theta))}{d\theta}=(-r\sin\theta , r\cos\theta)$
Similarly, the vector tangent to the line at $r$ is
$\vec{b}=\frac{d(x(r,\theta),y(r\theta))}{dr}=\frac{d(r\cos(\theta),(r\sin(\theta))}{dr}=(\cos \theta , \sin\theta)$
The Dot Product.
$\vec{a} \cdot \vec{b}= |\vec{a}| |\vec{b}| \cos(\phi)$ where $\phi$ is the angle between $\vec{a}$ and $\vec{b}$.
$\implies$
$\vec{a} \cdot \vec{b}=0 \rightarrow \vec{a} \perp \vec{b}$.
$\vec{a} \cdot \vec{b} = -r \sin \theta \cos\theta + r \cos \theta \sin \theta=0$
Therefore, the vectors are orthogonal.
Assume the line is crossing the unit circle where $\theta=0$ then
$a=(0,1)|_{\text{where the line crosses}}$ and $b=(1,0)|_{\text{where the line crosses}}$.
And you can always define any circle to be a unit circle by changing your units of measurement. So this particular choice of $a$ and $b$ really holds for any circle.