4. Given in the complex plane, the vertices $A, B, C$ of $\triangle A B C$ correspond to the complex numbers $3+2 \mathrm{i}, 3 \mathrm{i}, 2-\mathrm{i}$, respectively. The moving point $P$ corresponds to the complex number $z$. If the equation $|\bar{z}|^{2}+\alpha z+\bar{\alpha} \bar{z}+\beta=0$ represents the circumcircle of $\triangle A B C$, find the complex numbers $\alpha, \beta$.
4. It is known that $|AB|=|AC|$ and $|AB|^2+|AC|^2=|BC|^2$, so $\triangle ABC$ is an isosceles right triangle.

Let $z=x+y \mathrm{i} (x, y \in \mathbf{R})$, then the equation of the circle with the hypotenuse $BC$ as the diameter is $x(x-2)+(y+1)(y-3)=0$, i.e., $x^2+y^2-2x-2y-3=0$. Let $\alpha=a+b \mathrm{i}, \beta=c+d \mathrm{i} (a, b, c, d \in \mathbf{R})$, substituting into the known equation, we get $x^2 + y^2 + 2ax - 2by + c + d \mathrm{i} = 0$. Thus, we have $d=0$ and $x^2 + y^2 + 2ax - 2by + c = 0$. Comparing coefficients, we get $a=-1, b=1, c=-3$. Therefore, $\alpha=-1+\mathrm{i}, \beta=-3$.