Since we can fill a circle with infinitely-many copies of its radius, isn't $\frac{\text{Circle area}} {\text{Circle radius}} =\infty?$

You are approximating the circle with the sum of the areas of triangles that fit inside. This is actually the first numerical way pi was calculated, done by Archimedes about 2300 years ago. Yes you can make the triangular slices thinner and thinner, in a process where you can fit in triangles without bound, however because they are getting thinner in proportion with the rate at which you are making more of them, the sum of the areas of the triangles actually stays finite.

You do not fill the circle with copies of its radii (which will only have linear length and not width and so will not have any area). You fill the circle with tiny triangle slivers whose altitudes are nearly the same of the height and whose bases are really really small.

And you don't fill the circle with infinitely many. You fill the triangle with a lot of them but always so finite number even though it could be very large. And you reason what the area becomes when you take more and more of them.

And we don't fill the circle with them we almost fill the circle with them and we estimate.

SO for example if we fill in with $n$, a huge number, of triangle slivers, the height of on of this slivers will be nearly equal to the radius $r$ (just a teeny bit less) and base of these triangles, since there are $n$ of them and they are fitted along the circumference of the circle will have a $b$ very nearly equal to the the circumference divided by the number of triangles, that is $\frac Cn$.

So a triangle wedge has base = $b= \frac Cn$ and has a height $h = r$. So the area of the wedge is $\frac 12\cdot \frac Cn \cdot r$. And there are $n$ of them so the total area is $\require{cancel} n\cdot \frac 12 \cdot \frac Cn \cdot r = \cancel n\cdot \frac 12 \cdot \frac C{\cancel n}\cdot r = \frac 12 C\cdot r$. (Approximately. The larger $n$ is the closer this will be to the value and in a circle it will be reached.)

And $\frac {Area}{Radius} = \frac {\frac 12 Cr}{r} = \frac 12 C$.

Which considering $C = \pi \cdot diameter = 2\pi r$ is exactly what we would hope for.

Likewise the formula for the area is what we are used to if we consider $Area = \frac 12 C \cdot r = \frac 12 (\pi \cdot diameter)\cdot r= \frac 12(\pi \cdot 2r)\cdot r = \pi r^2$

You could just as easily claim that, if a square of side length $L$ is positioned with its sides horizontal and vertical, infinitely many line segments parallel to the leftmost side can fit in the square, and therefore its area is infinite rather than $L^2$. The problem in both arguments is line segments have zero area, and infinitely many zeroes don't have an infinite sum. Having said that, we also, of course, can't say the area is $0$ either. So what's going on here?

The reason a line segment has zero area is because every positive number is an upper bound on its area, because it can be made part of an arbitrarily thin rectangle. If the square is divided into $n$ strips of width $L/n$, each has area $L^2/n$, and the total is $L^2$. Admittedly, "using a rectangle's area to compute a square's area" may sound circular; but the area axioms include a rectangle's area, as well as the "subset has at most a superset's area" principle I used to bound a line segment's area.

Now let's divide a circle into $n$ sectors with $n$ equally spaced radii, so each sector subtends $2\pi/n$ radians. Each sector's vertices are those of a triangle that's a subset of the sector, so the sector has at least the triangle's area, $\tfrac12 r^2\sin(2\pi/n)$. So for all positive integers $n$, the circle's area is at least $\tfrac12nr^2\sin(2\pi/n)$. But this doesn't grow in proportion to $n$, the way an "it contains infinitely many radii so has infinite area" intuition may expect. This approaches $\pi r^2$ as $n\to\infty$. Or you can use the OP's diagram's "rearrange them into an approximate rectangle argument, thereby avoiding the above trigonometric limit.