Sangaku: Show line segment is perpendicular to diameter of container circle

Let $A$ be the big circle with centre $O$ and diameter $PQ$
Let $B$ be the circle internally tangent to $A$ at $P$ and intersecting $PQ$ again at $M$
Let $R$ be on $A$ such that $\overline{MR} = \overline{QR}$
Let $C$ be the circle that is tangent to $A$, $B$, $MR$ and inside $A$ and outside both $B$ and completely on the same side of $PQ$ as $R$
Invert at $M$, mapping $P$ to $Q$
Then $A$ maps to $A$, $B$ maps to the line $T$ that is tangent to $B$ at $Q$, $MR$ maps to $MR$, and thus $C$ maps to the circle $D$ that is tangent to $A$, $T$, $MR$ and outside $A$ and on the same side of $T$ as $M$ and completely on the other side of $PQ$ as $R$
Let $K$ be the point on $MQ$ such that $RK \perp MQ$
Let $N$ be the point such that $PQ \perp MN$ and $\overline{MN} = 2\overline{MR}$ and $N$, $R$ are on the opposite side of $PQ$
Let $X$ be the intersection of $NO$ and $A$ between $N$ and $O$
Let $Y$ be the point on $T$ such that $NY \perp T$
Let $Z$ be the point on $MR$ such that $NZ \perp MR$
Let $\overrightarrow{MO} = r \overrightarrow{OQ}$ and WLOG $\overline{OQ} = 1$
Then $\overline{MR}^2 = \overline{RK}^2 + \overline{MK}^2 = \overline{OR}^2 - \overline{OK}^2 + \overline{MK}^2 = 1 - (\frac{1-r}{2})^2 + (\frac{1+r}{2})^2 = 1+r$
Thus $\overline{NO}^2 = \overline{MN}^2 + r^2 = 4 \overline{MR}^2 + r^2 = 4+4r+r^2 = (2+r)^2$
Thus $\overline{NX} = (2+r)-1 = \overline{MQ}$
Also it is clear that $\overline{NY} = \overline{MQ}$
Also since $\triangle MNZ \sim \triangle RMK$, $\overline{NZ} = 2 \overline{MK} = \overline{MQ}$
Since $D$ is uniquely defined and $N$ is the centre of an identically defined circle, $N$ is the centre of $D$
Since the centres of $C$ and $D$ are collinear with $M$, the centre of $C$ lies on $MN$, therefore the line through both $M$ and the centre of $C$ is perpendicular to $PQ$
(QED)

Note that the same solution applies to both cases where $C$ is on either side of $PQ$, with very minor changes. Inversion seems to simplify most of these kind of questions. Quite a few can be found online.

For those who do not like inversion... =P

Let $A$ be the big circle with centre $O$ and diameter $PQ$
Let $B$ be the circle internally tangent to $A$ at $P$ and intersecting $PQ$ again at $M$ and having centre $J$
Let $R$ be on $A$ such that $\overline{MR} = \overline{QR}$
Let $C$ be the circle that is tangent to $A$, $B$, $MR$ and inside $A$ and outside both $B$ and completely on the same side of $PQ$ as $R$
Let $K$ be the point on $MQ$ such that $RK \perp MQ$
Let $D$ be the circle tangent to $B$ and $MR$ with centre $N$ such that $PQ \perp MN$ and $N$,$R$ are on the same side of $PQ$
Let $Z$ be the point where $D$ is tangent to $MR$
Let $\overrightarrow{MO} = r \overrightarrow{OQ}$ and WLOG $\overline{OQ} = 1$
Let $x = \overline{MN}$
Let $y$ be the radius of $D$
Then $\overline{MR}^2 = \overline{RK}^2 + \overline{MK}^2 = \overline{OR}^2 - \overline{OK}^2 + \overline{MK}^2 = 1 - (\frac{1-r}{2})^2 + (\frac{1+r}{2})^2 = 1+r$
Since $\triangle MNZ \sim \triangle RMK$, $\frac{x}{y} = \overline{MN} / \overline{NZ} = \overline{RM} / \overline{MK} = \sqrt{1+r} / \frac{1+r}{2} = \frac{2}{\sqrt{1+r}}$
Thus $(1+r)x^2 = 4y^2$
Also $(y+\frac{1-r}{2})^2 = \overline{NJ}^2 = \overline{MN}^2 + \overline{MJ}^2 = (\frac{1-r}{2})^2 + x^2$
Thus $(1+r)y^2 + (1+r)(1-r)y = (1+r)x^2 = 4y^2$
Since $y\not=0$, $(3-r)y = 1-r^2$
$\overline{OQ} = y + \overline{ON}$
$\Leftrightarrow (1-y)^2 = \overline{ON}^2 = \overline{MN}^2 + \overline{MO}^2 = x^2 + r^2$
$\Leftrightarrow (1+r)(1-2y+y^2) = 4y^2 + (1+r)r^2$
$\Leftrightarrow (3-r)y^2 + (2+2r)y - (1+r)(1-r^2) = 0$
$\Leftrightarrow (1-r^2)y + (2+2r)y = (1+r)(1-r^2)$
$\Leftrightarrow (1+r)(3-r)y = (1+r)(1-r^2)$ which is clearly true
Thus $D$ is tangent to $A$
Since $C$ is uniquely defined and $D$ is an identically defined circle, $N$ is the centre of $C$
Therefore the line through both $M$ and the centre of $C$ is perpendicular to $PQ$
(QED)