When rays of light that are parallel to the principal axis in a concave mirror reflects it will be reflected?

I was trying to prove that during reflection from a concave mirror, all rays parallel to the principal axis will meet the principal axis at a point (focus) that is equidistant from the centre of curvature and the pole.

I thought of proving this using coordinate geometry (and some trigonometry). I thought of proving it by proving that for a ray at an arbitrary height from the principal axis, parallel to it, the reflected ray will intersect the principal axis at a distance of $\dfrac{R}{2}$ from the centre of curvature, where $R$ is the radius of curvature of the concave mirror.

So, I plotted a concave mirror on a Cartesian Plane using the equation $x^2+y^2=R^2$ and also added a little constraint to make it look more like a concave mirror. The constraint was $x > 0$, $-R+1 \leq y \leq R-1$, but that's not the point here. Here's what the mirror looked like :

Now, I assumed that the incident ray is parallel to the principal axis at a height of $h$ from it. So, it's equation becomes $y=h$. Let's take values of $h$ such that $0 \leq h \leq R -1.5$ for the sake of simplicity here. Now, to find the point of incidence of the ray on the mirror, I found out the point of intersection of the curves of the two equations obtained above. As the point of incidence is a part of the incident ray, so, it's $y$ coordinate is $h$. Let it's $x$ coordinate be $a$. So, $a^2+h^2=R^2 \implies a = \sqrt{R^2-h^2}$. Hence, the point of incidence becomes : $(\sqrt{R^2-h^2},h)$.
Here's what the graph looked then :

$\Big($Note : I have added a little constraint to $y=h$ to limit it till the mirror only, the constraint is $x \leq \sqrt{R^2-h^2}\Big )$

Now, I made the tangent at the point of incidence and the normal too. The tangent was made using the equation $y=\dfrac{-ax+a^2+b^2}{b}$ where $(a,b)$ is the point where the tangent touches the circle. In this case, $a = \sqrt{R^2-h^2}$, $b = h$, so $y = \dfrac{-x\sqrt{R^2-h^2}+R^2}{h}$ becomes the equation for the tangent. For the normal, I just joined $(0,0)$ and $(\sqrt{R^2-h^2},h)$.
Here's what I obtained :

Now, I thought that if I could find another point on the reflected ray, then I would have two points, that point that I will obtain and the point of incidence. Using these two points, I can figure out the equation of the reflected ray. Now, to find the other point I took a point $(0,h)$ on the incident ray and then evaluated what its coordinates on the reflected ray will be. I observed the behavior of the point when it will be rotated with the point of incidence as it's centre and the angle of rotation equal to $2\alpha$, where $\alpha$ is the angle of incidence, that is, the angle between the incident ray and the normal.
The following diagram would help to explain it better :

Now, evaluating this was lengthy. I made another coordinate system, where the origin was $(\sqrt{R^2-h^2},h)$. Let's say that $f_2$ is a function that converts the coordinates of a point from the first Cartesian plane to the second. So, $f_2(x,y) = (x-\sqrt{R^2-h^2}),y-h)$. This converts the point $(0,h)$ to $(0-\sqrt{R^2-h^2},h-h) = (-\sqrt{R^2-h^2},0)$
The new Cartesian plane is shown in the diagram below. I have also drawn a dotted circle with radius equal to $\sqrt{R^2-h^2}$.

Now, in the Cartesian Plane, the coordinates of $(m,n)$ are $(r\cos(\pi+2\alpha),r\sin(\pi+2\alpha))$ $(\pi = \pi^c = \pi \text{ rads} = 180^o)$, and $r$ is the radius of the circle, which means that $r = \sqrt{R^2-h^2}$
Using $\sin(\pi+\theta) = -\sin\theta$ and $\cos(\pi+\theta) = -\cos\theta$, we get that the coordinates of $(m,n)$ in the second plane are $(-\sqrt{R^2-h^2}\cos (2\alpha), -\sqrt{R^2-h^2}\sin (2\alpha))$.

Now, let $f_2^{-1}$ be the function that converts the coordinates of a point from the second plane to the first. So, $f_2^{-1}(x,y) = (x+\sqrt{R^2-h^2},y+h)$.
So, $(m,n) = (\sqrt{R^2-h^2}-\sqrt{R^2-h^2}\cos(2\alpha), h - \sqrt{R^2-h^2}\sin(2\alpha))$.

Now, if we know the values of $\sin\alpha$ and $\cos\alpha$, we can use $\sin(2\theta) = 2\sin\theta\cos\theta$ and $\cos(2\theta) = \cos^2\theta - \sin^2\theta$, we can find the values of $\sin(2\alpha)$ and $\cos(2\alpha)$ in terms of $R$ and $h$.

Now, in Figure 4, $\alpha$ is the angle made by the origin, the point of incidence and $(0,h)$.
Using $\sin\theta = \dfrac{\text{Perpendicular side}}{\text{Hypotenuse}}$, we obtain : $\sin\alpha = \dfrac{h}{R}$. And as $\cos\theta = \sqrt{1-\sin^2\theta}$, we get : $\cos\alpha = \dfrac{\sqrt{R^2-h^2}}{R}$.

So, $\sin(2\alpha) = \dfrac{2h\sqrt{R^2-h^2}}{R^2}$, $\cos(2\alpha) = \dfrac{R^2-2h^2}{R^2}$ (with some calculations that I omitted because the question is long enough already).

Now, let the equation of the reflected ray be $y=px+c$. We know that $(m,n)$ and $(\sqrt{R^2-h^2},h)$ lie on the line. So, we can form two equations that are : $$h = p\sqrt{R^2-h^2}+c$$ $$h-\sqrt{R^2-h^2}\sin(2\alpha) = p\sqrt{R^2-h^2}-p\sqrt{R^2-h^2}\cos(2\alpha) + c$$ On subtracting the second equation from the first one, we obtain : $$\sqrt{R^2-h^2}\sin(2\alpha) = p\sqrt{R^2-h^2}\cos(2\alpha) \implies p = \tan(2\alpha)$$ $$c = h-p\sqrt{R^2-h^2} \implies c = h-\sqrt{R^2-h^2}\tan(2\alpha)$$ $$\tan(2\alpha) = \dfrac{\sin(2\alpha)}{\cos(2\alpha)} \implies \tan(2\alpha) = \dfrac{2h\sqrt{R^2-h^2}}{R^2-2h^2}$$ Finally, with some simplification, we obtain the equation of the reflected ray as : $$y = \dfrac{2hx\sqrt{R^2-h^2}-R^2h}{R^2-2h^2}$$

Now, everything seemed fine till I started changing the value of $h$ to see where the reflected ray will intersect the principal axis. The point was not the same every time. In fact, only when $h \rightarrow 0$, the point approached what's supposed to be the focus. In the above diagrams, I have taken $R = 10$, so, the reflected ray should have intersected the principal axis at $(5,0)$. But, there is clearly some deviation.

I thought of visualizing the deviation by comparing my reflected ray with the ray joining the point of incidence and the focus. Here's what it looked like :

I observed that in my case, the angle of incidence and reflection actually appear to be equal while that is not the case when the reflected ray passes through the focus. I also used the protractor in Desmos to check this.

Now, where did I go wrong? I have recalculated this thrice and I did not find a single error. The only possible error, in my opinion, is that I have assumed that the reflection from a concave mirror happens with the line joining the point of incidence and the centre of curvature as the normal at the point of incidence. On the contrary, I am pretty sure that this assumption is correct.

I would like to know why my reflected ray is not passing through the focus.
Thanks!

PS : Thank You for taking the time to read this question. It took me a long time to write it and so, I was unable to recheck it for errors. So, there might be a few of them and I would be more than happy if you correct them by editing this question or inform me about them, so that I can correct them.
PPS : Here's my graph in Desmos