Let $AP$ and $BQ$ intersect at $D$ and let $R$ and $S$ be the centres of the circles. Join $PR$, $QS$, $PC$, $QC$ and $DC$ and note the circles touch at $C$. Let $DC$ meet $PQ$ at $E$. Since $PR$ and $QS$ are both perpendicular to $PQ$, $PR$ is parallel to $QS$ and so $\angle PRA = \angle QSC$ which implies that triangles $PRA$ and $QSC$ are similar since both triangles are isosceles. It follows that $\angle PAC = \angle QCB$.

Since $\angle APC = 90^\circ$ (angle in semicircle), we have $90^\circ = \angle PAC + \angle PCA = \angle QCB + \angle PCA$, hence $\angle PCQ = 90^\circ$. Also $\angle CQB = 90^\circ$ (angle in semicircle).
Because $\angle DPC = \angle PCQ = \angle DQC = 90^\circ$, we also have $\angle PDQ = 90^\circ$, which shows that $D$ lies on the circumference of the circle on $AB$ as diameter. In addition $PCQD$ is a rectangle and since the diagonals of a rectangle bisect each other $|EP| = |EQ|$, hence $E$ lies on the radical axis of the two circles and so does $C$. This means that $EC$, and so $DC$ as well, is the common tangent at $C$. This completes the proof.

Let $D$ be the intersection point of the lines $AP$ and $BQ$.
Because $PQ$ is a tangent to the circle with diameter $AC$, the Alternate Segment Theorem gives $\angle QPC = \angle PAC$. The angle $\angle DPC$ is a right angle, because $\angle APC$ is standing on a diameter. Hence, $\angle DPQ = 90^\circ - \angle QPC = 90^\circ - \angle PAC$.
Working with the circle of diameter $CB$ we obtain in a similar way $\angle PQC = \angle QBC$ and $\angle DQP = 90^\circ - \angle PQC = 90^\circ - \angle QBC$.

Considering triangle $DPQ$ we now obtain
$$\begin{aligned} \angle PDQ &= 180^\circ - (\angle DPQ + \angle DQP) \\ &= 180^\circ - (90^\circ - \angle PAC + 90^\circ - \angle QBC) \\ &= \angle PAC + \angle QBC. \end{aligned}$$
Using the angle sum in triangle $DAB$ this implies $2\angle PDQ = 180^\circ$, i.e. $\angle PDQ$ is a right angle. This shows that $D$ is on the circle that has $AB$ as a diameter. This also shows that $PCQD$ is a rectangle, which implies that $\angle PDC = \angle PQC$.
Together with $\angle PQC = \angle QBC$, which we have seen above, we now obtain
$$\angle PAC + \angle PDC = \angle PAC + \angle QBC = \angle PDQ = 90^\circ.$$
From triangle $DAC$ we now see that $\angle ACD = 90^\circ$, hence $D$ is on the common tangent to both circles at $C$.

Let $D$ be the intersection point of the lines $AP$ and $BQ$.
The angles $\angle APC$ and $\angle CQB$ are standing on diameters, hence these are right angles. Therefore, the quadrilateral $PCQD$ is cyclic and so
$$\angle PDC = \angle PQC \quad \text{and} \quad \angle QPC = \angle QDC.$$

Because $PQ$ is a tangent to the circle with diameter $AC$, the Alternate Segment Theorem gives $\angle QPC = \angle PAC$. Similarly, $\angle PQC = \angle QBC$. Combined with the equations above, we obtain
$$\angle PDC = \angle QBC \quad \text{and} \quad \angle QDC = \angle PAC$$
which shows that the triangles $ADC$, $DBC$ and $ABD$ are similar. As a consequence, $\angle DCA = \angle DCB = \angle BDA$. Because $\angle DCA + \angle DCB = 180^\circ$, these three angles are right angles. This implies that $D$ is on the circle with diameter $AB$ and $D$ is on the common tangent to both circles at $C$.

Let $D$ be the intersection point of the line $AP$ and the common tangent at $C$ of the circles with diameters $AC$ and $CB$. Let $N$ be the intersection point of $PQ$ and $CD$. Because tangents from a point to a circle are of equal length, we have $|NP| = |NC|$ and $|NC| = |NQ|$, hence $N$ is the circumcentre of triangle $PCQ$. As $N$ is on $PQ$, $\angle PCQ$ is a right angle. This implies that $\angle QPC + \angle PQC = 90^\circ$.
The angles $\angle APC$ and $\angle CQB$ are standing on diameters, hence these are right angles. Therefore, $\angle PAC + \angle PCA = \angle QCB + \angle QBC = 90^\circ$.
Because $PQ$ is a tangent to the circles with diameters $AC$ and $CB$, the Alternate Segment Theorem gives $\angle QPC = \angle PAC$ and $\angle PQC = \angle QBC$. Together with equations obtained earlier, we now have $\angle PAC = \angle QPC = \angle QCB$ and $\angle QBC = \angle PQC = \angle PCA$.
Since $CD$ is perpendicular to $AB$, we also have $\angle PAC + \angle PDC = 90^\circ$ from which we obtain $\angle PDC = \angle PQC$. This equality implies that $D$ is on the circumcircle of triangle $PCQ$. So, $D$ is the intersection point of the common tangent at $C$ and the circumcircle of triangle $PCQ$.

We now let $E$ be the intersection point of the line $BQ$ and the common tangent at $C$ of the circles with diameters $AC$ and $CB$. We do not know at this stage if $E = D$, as indicated in the diagram.
From the equations obtained above and $\angle QBC + \angle QEC = 90^\circ$ we obtain $\angle QEC = \angle QPC$. This equality implies that $E$ is on the circumcircle of triangle $PCQ$. So, $E$ is the intersection point of the common tangent at $C$ and the circumcircle of triangle $PCQ$, which means that $E = D$. This proves that $AP$, $BQ$ and the common tangent at $C$ are concurrent.
To finish, we just need to note that $\angle ADB = \angle PDC + \angle QEC = 90^\circ$ from the equations obtained before, and this implies that the point $D = E$ is on the circle with diameter $AB$.

Let $D$ be one of the intersection points of the circle with diameter $AB$ and the common tangent at $C$ of the circles with diameters $AC$ and $CB$. Connect $D$ to $A$ and let $P'$ be the intersection point of this line with the circle that has diameter $AC$. Connect $D$ to $B$ and let $Q'$ be the intersection point of this line with the circle that has diameter $CB$. We wish to prove that $P' = P$ and $Q' = Q$. This follows when we have shown that $P'Q'$ is a common tangent to the circles with diameters $AC$ and $CB$.

To prove this, we first notice that, since $\angle AP'C$, $\angle CQ'B$ and $\angle ADB$ are right angles, we have $\angle DAC = \angle Q'CB$ and $\angle DBC = \angle P'CA$.
Because $CD$ is the radical axis of the circles with diameters $AC$ and $CB$, the point $D$ has the same power with respect to these two circles, that is $|DP'| \cdot |DA| = |CD|^2 = |DQ'| \cdot |DB|$ and so
$$\frac{|DA|}{|DB|} = \frac{|DQ'|}{|DP'|}$$
which implies that triangles $DAB$ and $DQ'P'$ are similar in such a way that
$$\angle DBC = \angle DP'Q' \quad \text{and} \quad \angle DAC = \angle DQ'P'.$$
Using the initially noticed equalities, we see now that
$$\angle P'CA = \angle DP'Q' \quad \text{and} \quad \angle Q'CB = \angle DQ'P'.$$
The converse of the Alternate Segment Theorem implies now that $P'Q'$ is tangent to the circles with diameters $AC$ and $CB$. This shows that $P' = P$ and $Q' = Q$ and finishes the proof.