Using Radical Axis to Prove Concurrency A, C, B, D are four collinear points. The circle with diameter $AB$ intersects the circle of diameter $CD$ at $M$ and $N$. Let $P$ be a point on the line $MN$ such that $M$ is between $N$ and $P$. Segment $PB$ intersects the circle of diameter $AB$ at $R$ and $B$, while the segment $PC$ intersects the circle of diameter $CD$ at $Q$ and $C$. Show that the lines $MN, AR, DQ$ are concurrent. 
 A: Let $O$ be the intersecting point of $AD$ and $PN$. Let $AR$ meet $PN$ at $X_1$, and let $DQ$ meet $PN$ at $X_2$. First, notice that $$PQ\cdot PC = PM\cdot PN = PR\cdot PB$$
where we look at the power of point $P$ to the right and left circles respectively. Then, notice that $$PN\perp AD\ \text{and} \ \angle ARB = 90^\circ \implies X_1,R,B,O \ \ \text{are cyclic} \implies PR\cdot PB = PX_1\cdot PO$$$$PN\perp AD\ \text{and} \ \angle DQC = 90^\circ \implies X_2,Q,C,O \ \ \text{are cyclic} \implies PQ\cdot PC = PX_2\cdot PO$$
Combine these three equations to get $$ PX_1\cdot PO=PR\cdot PB =PQ\cdot PC =PX_2\cdot PO\implies PX_1=PX_2\implies X_1=X_2$$ Hence the three lines $MN, AR, DQ$ are concurrent as desired.
A: A proof with inversion. Draw a tangent $PT$ to one of the circles, say the left one, where $T$ is the point of tangency, and draw the circle $k_P$ with center $P$ and radius $PT$. Since $MN$ is the radical axis of the two circles and $P \in MN$, the circle $k_P$ is orthogonal to both circles. Draw circle $k_1$ circumscribed around triangle $PQR$. The inversion with respect to circle $k_P$ maps $k_1$ to the straight line $AB$ and $Q$ and $R$ are mapped to $C$ and $B$ respectively. The line $MN$ passes through the center of inversion $P$ and since it is orthogonal to $AB$, it must be orthogonal to the preimage circle $k_1$. Let $H$ be the second intersection point of $k_1$ with $MN$, the first one being $P$. Then $PH$, being orthogonal to $k_1$ is its diameter so $\angle \, PRH = \angle \, PQR = 90^{\circ}$ i.e. $SR \perp BP$ and $HR \perp CP$. However, since $AB$ is a diameter of the first circle, $AR \perp BP$. Due to the fact that there is a unique line through point $R$ perpendicular to $BP$, the two lines $HR$ and $AR$ coincide so $AR$ passes through $H$. Analogously, since $CD$ is a diameter of the second circle, $DQ \perp CP$. Due to the fact that there is a unique line through point $Q$ perpendicular to $CP$, the two lines $HQ$ and $DQ$ coincide so $DQ$ passes through $H$. Consequently, the three lines $MN, \, AR, \, DQ$ pass through the common point $H$.            
