4. In the Cartesian coordinate system, points $O(0,0), A(0,1), B(2,0)$, and $C(4,0)$ are given. Find all possible coordinates of point $M$ such that

$$
\measuredangle O M A=\measuredangle C M B \quad \text { and } \quad \measuredangle M A O=\measuredangle M B C .
$$
4. Let $(x, y)$ be the coordinates of point $M$. From the condition of the problem, it follows that $\triangle M A O \sim \triangle M B C$, so we get $\frac{M C}{M O}=\frac{M B}{M A}=\frac{B C}{A O}$, which (after squaring) reduces to $\frac{(4-x)^{2}+y^{2}}{x^{2}+y^{2}}=\frac{(2-x)^{2}+y^{2}}{x^{2}+(y-1)^{2}}=4$, and from here we have the system $16-8 x+x^{2}+y^{2}=$ $4 x^{2}+4 y^{2}$ and $4-4 x+x^{2}+y^{2}=4 x^{2}+4 y^{2}-8 y+4$. Subtracting the first equation from the second gives $4 x-12=-8 y+4$, i.e., $x=4-2 y$. Substituting this, for example, into the second equation, we get $4-4(4-2 y)+(4-2 y)^{2}+y^{2}=4(4-2 y)^{2}+4 y^{2}-8 y+4$, which after simplification gives $-15 y^{2}+64 y-64=0$. Solving this equation, we get $y_{1 / 2}=$ $\frac{-64 \pm \sqrt{64^{2}-4 \cdot 15 \cdot 64}}{-30}=\frac{-64+16}{-30}$, i.e., $y_{1}=\frac{8}{5}$ and $y_{2}=\frac{8}{3}$. The corresponding values for $x$ are $x_{1}=4-\frac{16}{5}=\frac{4}{5}$ and $x_{2}=4-\frac{16}{3}=-\frac{4}{3}$.

Therefore, there are two possible coordinates for point $M: \left(\frac{4}{5}, \frac{8}{5}\right)$ and $\left(-\frac{4}{3}, \frac{8}{3}\right)$.