Answer: $a = b = 2$.
Let $A(\alpha, 0)$, $B(\beta, 0)$, $C(0, \gamma)$, and $M(a, b)$. Since $M$ is the center of the circumscribed circle of the triangle $ABC$, $M$ lies on the perpendicular bisector of the segment $AB$. Therefore, the projection of $M$ on the $x$-axis is the midpoint of the segment $AB$, i.e. $a = (\alpha + \beta)/2$.
Since the points $A$ and $B$ lie on the parabola, their abscissae satisfy the equation $x^2 - 2bx + a + 1 = 0$, then, by Vieta's theorem, $a = (\alpha + \beta)/2 = 2b/2 = b$.
Substituting $x = 0$ into the parabola equation, we find the ordinate $\gamma$ of $C$: $\gamma = a + 1$.
Let $R$ be the radius of the circle $\Gamma$. Since $C$ lies on $\Gamma$, we obtain
$$R^2 = |MC|^2 = (a-0)^2 + (b-\gamma)^2 = [a=b, \gamma=a+1] = a^2+1.$$
Since $A$ and $B$ lie on $\Gamma$, we get
$$a^2 + 1 = R^2 = |MA|^2 = |MB|^2 \Rightarrow (a - \alpha)^2 + (b - 0)^2 =$$
$$= (a - \beta)^2 + (b - 0)^2 = (a - \alpha)^2 + a^2 = (a - \beta)^2 + a^2 \Rightarrow (a - \alpha)^2 = (a - \beta)^2 = 1.$$
Without loss of generality we can assume that $\beta > \alpha$, then $\alpha = a - 1$, $\beta = a + 1$. The numbers $\alpha = a - 1$ and $\beta = a + 1$ are the roots of the equation $x^2 - 2bx + a + 1 = 0$, so, by Vieta's theorem
$$a^2 - 1 = (a - 1)(a + 1) = \alpha\beta = a + 1 \Rightarrow a^2 - a - 2 = 0.$$
We see that $a = -1$ and $a = 2$ are the roots of this equation.

For $a = -1$ we have $\gamma = 0$, but then $C(0, 0)$, i.e. $C$ coincides with the origin, contrary to the problem condition.
For $a = 2$ we have $M(2, 2)$, the parabola equation has the form $y = x^2 - 4x + 3$, and the circle equation has the form $(x - 2)^2 + (y - 2)^2 = 5$. It is easy to verify that all problem conditions hold in this case.