Description of the construction. Let $O_P$ and $O_Q$ denote the centres of the circles $\Omega_P$ and $\Omega_Q$, respectively. Draw the circle centre $A$ that passes through $B$ and let $K$ and $L$ be its intersection points with $\Omega_P$ and $\Omega_Q$, respectively. Next construct the angle bisector of $\angle KBL$, then the perpendicular to this angle bisector that passes through $A$. Let $P$ and $Q$ be the points where this perpendicular intersects $\Omega_P$ and $\Omega_Q$, respectively. Let $M$ be the point where the lines $PO_P$ and $QO_Q$ meet. The circle centre $M$ that passes through $P$ is the required circle.

Justification of the construction. Let $\omega$ be the circle with centre $A$ that passes through $B$. When we invert the two given circles in $\omega$, we obtain the lines $BK$ and $BL$, where $K$ and $L$ are the intersection points of $\omega$ with $\Omega_P$ and $\Omega_Q$, respectively. The required circle inverts to a circle which contains $A$ and which is tangent to $BK$ and $BL$. The points of contact of the inverted circle with $BK$ and $BL$, say $P'$ and $Q'$, are the inverses of $P$ and $Q$, respectively. Because $A$ is collinear with every pair of inverse points, $P, A, Q$ are collinear exactly when $P', A, Q'$ are collinear. In this case, $P, P', A, Q, Q'$ are collinear. Finally, because $P'Q'$ is perpendicular to the angle bisector of angle $\angle KBL$, the points $P$ and $Q$ must be on the perpendicular to this angle bisector that passes through $A$.