Given a specific incident ray and a specific reflected ray, how to find the exact point of incidence on a circular convex mirror? In this illustration, the circle whose center is O is a perfectly circular convex mirror, AC is the incident ray and BC is the reflected ray. If the mirror and the points A and B are known, is there a way to find the exact location of the point of incidence C on the mirror?

 A: 
According to Fermat's principle of least time this problem is equivalent to find the shortest path $AC+BC=\texttt{minimum}$. In Figure above I give a graphical solution for the shortest path.
$\boldsymbol{\S}\:$A. Main Section
According to Fermat's principle of least time this problem is equivalent to finding the shortest path : $ \texttt{AC+CB}\boldsymbol{=}\texttt{minimum}$. This condition will ensure the two angles equality.
The idea is to provide an answer on the ground of the relevant problem in the case of a flat mirror. In the latter we take  the point $\,\texttt A'\,$ symmetric of the point $\,\texttt A\,$  with respect to the flat mirror line, joint points $\,\texttt{A,B}\,$  with a line that intersects mirror line at  point $\,\texttt C$, see Figure-01. Then $\texttt{AC+CB}\boldsymbol{=}\texttt{minimum}$ in the sense that  $\texttt{AC+CB}\boldsymbol{\le}\texttt A \texttt C'\texttt + \texttt C' \texttt B$ for any point $\,\texttt C'\,$ on the flat mirror.

The case of a circular convex mirror, see Figure-02, is not so simple. The infinitesimal arc shown could be considered as a flat mirror with respect to which the point $\,\texttt A'\,$ is the image of the fixed point $\,\texttt A$.

The set of all  images $\,\texttt A'\,$  of the fixed point $\,\texttt A\,$, a geometric locus, is a closed curve as shown in Figure-03. The properties of this curve will provide us later with the solution to the problem.

At first with the configuration shown the parametric equation of this closed curve is
At first with the configuration shown the parametric equation of this closed curve is
\begin{align}
\mathbf x\left(\phi\right)&\boldsymbol{=}
\begin{bmatrix}
x_1\left(\phi\right)\vphantom{\dfrac{a}{b}}\\
x_2\left(\phi\right)\vphantom{\dfrac{a}{b}}
\end{bmatrix}\boldsymbol{=}
\begin{bmatrix}
2R\cos\phi\boldsymbol{-}a\cos2\phi\vphantom{\dfrac{a}{b}}\\
2R\sin\phi\boldsymbol{-}a\sin2\phi\vphantom{\dfrac{a}{b}}
\end{bmatrix}\,,\quad \phi \in \left[\boldsymbol{-}\omega_{\texttt A},\boldsymbol{+}\omega_{\texttt A} \right]
\tag{01a}\label{01a}\\
\texttt{where} \quad \omega_{\texttt A} &\boldsymbol{=}\arccos{\left(R/a\right)}\in  \left[0,\pi/2\right]
\tag{01b}\label{01b}
\end{align}
and $\,R\boldsymbol{=}\text{the mirror radius}$,$\,a\boldsymbol{=}\text{distance of the fixed point $\,\texttt A\,$ from the mirror center}$. The parameter of this curve representation is the angle  $\,\phi\,$ that is the $''$ angular distance$''$ of the point of incidence from the $\,x_1\boldsymbol{-}$axis, see Figures-02 and -03.
A vector tangent to the curve of images at the image point $\,\texttt A'\,$ is
\begin{equation}
\mathbf t_{\texttt A'}\boldsymbol{\equiv}\dfrac{\mathrm d\mathbf x}{\mathrm d\phi}\boldsymbol{=}
\begin{bmatrix}
\mathrm dx_1/\mathrm d\phi\vphantom{\dfrac{a}{b}}\\
\mathrm dx_2/\mathrm d\phi\vphantom{\dfrac{a}{b}}
\end{bmatrix}\boldsymbol{=}2
\begin{bmatrix}
\boldsymbol{-}R\sin\phi\boldsymbol{+}a\sin2\phi\vphantom{\dfrac{a}{b}}\\
\hphantom{\boldsymbol{-}}R\cos\phi\boldsymbol{-}a\cos2\phi\vphantom{\dfrac{a}{b}}
\end{bmatrix}\qquad \texttt{(tangent vector)} 
\tag{02}\label{02}
\end{equation}
so a vector normal to the curve of images at the image point $\,\texttt A'\,$ is
\begin{equation}
\mathbf n_{\texttt A'}\boldsymbol{\equiv}
\begin{bmatrix}
R\cos\phi\boldsymbol{-}a\cos2\phi\vphantom{\dfrac{a}{b}}\\
R\sin\phi\boldsymbol{-}a\sin2\phi\vphantom{\dfrac{a}{b}}
\end{bmatrix}\qquad \texttt{(normal vector)} 
\tag{03}\label{03}
\end{equation}
(Note : The vector $\mathbf t_{\texttt A'}$ must not be confused with the unit tangent vector $\mathbf t$ from the theory of curves. These two vectors differ by a scalar factor. Also the vector $\mathbf n_{\texttt A'}$ must not be confused with the unit normal vector $\mathbf n$ or the curvature vector $\mathbf k$ from the theory of curves. These three vectors are collinear differing by scalar factors.)
Now the  vector $\,\mathbf n_{\texttt A'}\,$ is collinear to the incident ray $\,\texttt C\texttt A'\,$ (more precisely is exactly equal to the vector $\,\overset{\boldsymbol{\rightarrow}}{\texttt C\texttt A'}\,$ in Figure-03) so we have the following result

If an  incident ray is reflected through the fixed point $\,\texttt A\,$ then this ray is normal to the curve of images of this point  at its  image point  $\,\texttt A'$.


Consider now that $\mathbf b \boldsymbol{=}b\left(\cos\beta,\sin\beta\right)$ is the position vector of the other point $\,\texttt B\,$ as shown in Figure-04. To find the point of incidence $\,\texttt C\,$ we'll find an implicit equation for the angle $\,\phi\,$ by requiring the vector $\,\overset{\boldsymbol{\rightarrow}}{\texttt B\texttt A'}\boldsymbol{=}\mathbf x\left(\phi\right)\boldsymbol{-}\mathbf b\,$ to be normal to the tangent vector $\mathbf t_{\texttt A'}$ :
\begin{equation}
\langle \overset{\boldsymbol{\rightarrow}}{\texttt B\texttt A'},\mathbf t_{\texttt A'}\rangle\boldsymbol{=}\langle \mathbf x\left(\phi\right)\boldsymbol{-}\mathbf b,\mathbf t_{\texttt A'}\rangle\boldsymbol{=}0
\tag{04}\label{04}
\end{equation}
Inserting for $\,\mathbf x\left(\phi\right),\mathbf t_{\texttt A'}\,$ their expressions \eqref{01a},\eqref{02} respectively above inner product yields
\begin{equation}
\boxed{\:\:\dfrac{R}{b}\sin\phi\boldsymbol{+}\dfrac{R}{a}\sin\left(\phi\boldsymbol{-}\beta\right)\boldsymbol{-}\sin\left(2\phi\boldsymbol{-}\beta\right)\boldsymbol{=}0 \vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\:\:}
\tag{05}\label{05}
\end{equation}
Since
\begin{equation}
\dfrac{R}{a}\boldsymbol{=}\cos\omega_{\texttt A}\,,\qquad \dfrac{R}{b}\boldsymbol{=}\cos\omega_{\texttt B}
\tag{06}\label{06}
\end{equation}
the angles $\,\omega_{\texttt A},\omega_{\texttt B}\,$ shown in Figure-03 and Figure-04 respectively we have finally
\begin{equation}
\cos\omega_{\texttt B}\sin\phi\boldsymbol{+}\cos\omega_{\texttt A}\sin\left(\phi\boldsymbol{-}\beta\right)\boldsymbol{-}\sin\left(2\phi\boldsymbol{-}\beta\right)\boldsymbol{=}0
\tag{07}\label{07}
\end{equation}
This implicit equation could be solved with respect to $\,\phi\,$ numerically or graphically.
As a first quick check of above equation we note that if $\,b\boldsymbol{=}a\,$ then because of symmetry the line $\,\texttt K\texttt C\,$ must be bisector of angle $\,\beta\,$ so $\,\phi\boldsymbol{=}\beta/2$. Indeed, in that case this is a solution of equation \eqref{07}.
By an other  analysis for the  path $\texttt{AC+CB}$ in Figure-04 to be the shortest the angle $\,\phi\,$ must satisfy the equation
\begin{equation}
\boxed{\:\:\underbrace{\dfrac{a\sin\phi}{\sqrt{a^2\boldsymbol{+}R^2\boldsymbol{-}2aR\cos\phi}}}_{\sin\theta_{\texttt A}}\boldsymbol{-}\underbrace{\dfrac{b\sin\left(\beta\boldsymbol{-}\phi\right)}{\sqrt{b^2\boldsymbol{+}R^2\boldsymbol{-}2bR\cos\left(\beta\boldsymbol{-}\phi\right)}}}_{\sin\theta_{\texttt B}}\boldsymbol{=}0 \vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\:\:}
\tag{08}\label{08}
\end{equation}
Equation \eqref{08} gives the same solution as  \eqref{05}.
$=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!$
Example A
Consider the arrangement of Figure-04 with the following data :
\begin{equation}
R\boldsymbol{=}2\,\qquad a\boldsymbol{=}3\,\qquad b\boldsymbol{=}4\,\qquad \beta\boldsymbol{=-}\pi/5\boldsymbol{=-}36^{\rm o}
\tag{A-01}\label{A-01}
\end{equation}
This arrangement is shown with precision in Figure-05.


A: The direction to aim the light ray, from A, so that it bounces off the circular mirror and reaches B can be specified by the angle from OA, $\alpha$, see diagram.
$p$, $q$, $r$ and $\theta$ are known and we want to find $\alpha$.
Sine rule on triangle OCA and then OCB, and using $\sin (180 -t) = \sin t$
$$\sin t = \frac{p}{r}\sin \alpha$$
$$\sin t = \frac{q}{r}\sin \beta$$

Let angle OAB be $q'$ and OBA be $p'$,  From triangle ABC
$(q'-\alpha)+(p'-\beta)+2t = 180$
from triangle OAB
$\theta +p' +q' = 180$
so
$2t = \alpha +\beta + \theta$
$$2\sin^{-1}(\frac{p}{r}\sin \alpha) = \alpha +  \sin^{-1}(\frac{p}{q}\sin\alpha) + \theta\tag1$$
and $\alpha$ is the only unknown in this equation.
Desmos the graphical calculator provides a plot of the left and right hand side in the link below.  The point where the curves cross gives $\alpha$.
The values in the link can be changed, so it can be used to find $\alpha$ for different circumstances.  Just click near the crossing point to see the answer.
https://www.desmos.com/calculator/pegrk1ogzc
From the diagram in the question and the screen being used $p$ is 5.1cm, $q$ is 8.5cm, $r$ is 3.2cm and $\theta $ is 0.40 radians.  $\alpha$ comes out as 0.244 radians or 14 degrees.
It's been tested by using points A(-3,0) and B(-2,6) and making this image using equation 1) and the Desmos link, to find the angle from OA and from OB that the rays should point.  Those lines crossed on the circle and the angle of reflection is equal to the angle of incidence.

similarly if B is (2,6)

A: To complete the previous answers, you can obtain very simply the equation given above by writing that the point of contact I with the mirror is such that the optical path is extremal. With the obvious notations of the figure:
$$L=AI+IB=\sqrt{{d_A}^2+R^2-2Rd_Acos(\varphi)}+\sqrt{{d_B}^2+R^2-2Rd_Bcos(\theta_B-\varphi)}$$
The extremality condition is $\frac{dL}{d\varphi}=0$ or:$$\frac{Rd_Asin(\varphi)}{\sqrt{{d_A}^2+R^2-2Rd_Acos(\varphi)}}-\frac{Rd_Bsin(\theta_B-\varphi)}{\sqrt{{d_B}^2+R^2-2Rd_Bcos(\theta_B-\varphi)}}=0$$
As indicated, you have to solve this equation in $ \varphi$ to find the position of I.
This equation can be written $$\frac{d_Asin(\varphi)}{AI}=\frac{d_Bsin(\theta_B-\varphi)}{IB}$$ and in this form, we immediately see that it amounts to write $sin(i)=\frac{d_Asin(\varphi)}{AI}=\frac{d_Bsin(\theta_B-\varphi)}{IB}=sin(r)$ : equality of the angles of incidence and reflection.
