Result related to 3 standard intersectiing circles I am in the middle of solving a problem and stuck at the following link:-
Below is a standard diagram showing three circles [namely, $\omega=(ABCD) \; (CDPG) \;and\; (BCGQ)$] intersecting each other so that DCQ and BCP are straight lines. Let M be the midpoint of PQ. MC produced cuts $\omega$ at R.

If PR cuts $\omega$ at L, prove that L, D, G are collinear. Or equivalently, if we assume that PR and GD produced meet at L, prove that L is a con-cyclic point of $\omega$.
 A: Lemma 1: Given a pair of circles $c_1$ and $c_2$ there is a unique special line $l_{12}$, called the radical axis of $c_1$ and $c_2$, with the property that the center of any circle that is simultaneously perpendicular to both circles $c_1$ and $c_2$ lies on $l_{12}$. Conversely, any circle whose center is on the radical axis $l_{12}$ and is perpendicular to one of the circles $c_1$ or $c_2$ is necessarily perpendicular to the other circle too. The radical axis $l_{12}$ is always perpendicular to the line connecting the centers of $c_1$ and $c_2$. Moreover, if $c_1$ and $c_2$ intersect, the radical axis $l_{12}$ passes through the intersection points of $c_1$ and $c_2$.
Lemma 2: Let $c_0$ be a circle with center $O_0$ and $c$ be another circle with center $O$ so that the center $O_0$ of $c_0$ lies on the circle $c$. Then the inversion of circle $c$ with respect to $c_0$ is a line $l$ perpendicular to the line $OO_0$ and $l$ is in fact the radical axis of the pair of circles $c_0$ and $c$.
Radical Axis Theorem: Given three circles, every pair of circles has a unique radical axis. Since there are three pairs of circles, there are three radical axes. These radical axes intersect in a common point.
Solution: Draw the circles $c_P$ and $c_Q$ such that they are centered at the points $P$ and $Q$ respectively and are perpendicular to the circle $\omega = c(ABCD)$.

Denote by $T_1$ and $T_2$ the intersection points of circle $c_P$ with $\omega$. Consider the three circles $\omega, \, c_P$ and $c(CDPG)$. They form three pairs and there is one radical axis for each pair. By Lemma 1 the radical axes are
$$l\big(\omega, \, c(CDPG)\big) = CD, \,\,\,\, l(\omega, \, c_P) = T_1T_2, \,\,\,\, l\big(c_P, \, c(CDPG)\big) = AB \,\,\, \text{ by Lemma 2}$$ Now, by the Radical Axis Theorem, the three radical axes $CD, \, T_1T_2$ and $AB$ intersect in a common point, which is point $Q$. This, in particular implies that $Q$ lies on the radical axis $T_1 T_2$.
Now, by Lemma 1, since the center $Q$ of $c_Q$ lies on the radical axis $T_1T_2$ and by construction $c_Q$ is perpendicular to $\omega$, the circle $c_Q$ is perpendicular to the circle $c_P$. This means that if $E_1$ and $E_2$ are the intersection points of $c_P$ and $c_Q$ then $\angle \, PE_1Q = PE_2Q = 90^{\circ}$. Therefore, if $c_M$ is the circle with center $M$ and diameter $PQ$, it passes also through the points $E_1$ and $E_2$. However, by Lemma 1, $E_1E_2$ is the radical axis of $c_P$ and $c_Q$ and again by Lemma 1, $E_1E_2$ is in fact the radical axis of, for example, $c_P$ and $c_M$. Thus, by Lemma 2, since $\omega$ is perpendicular to $c_P$ it must be also perpendicular to $c_M$.
Perform inversion in circle $c_M$. As $\omega$ is perpendicular to $c_M$, it is inverted into itself and moreover point $C$ is mapped to point $R$. By the properties of inversion
$$MC \cdot MR = MP^2$$ which translates into
$$\frac{MC}{MP} = \frac{MP}{MR}$$ which combined with $\angle \, CMP = \angle \, PMR$ shows that triangles $\Delta\,CMP$ and $\Delta \, PMR$ are similar and hence
$$\angle \, CPM = \angle \, PRM = \alpha$$ Furthermore, by assumption, quad $ABCR$ is inscribed in circle $\omega$, hence
$$\angle \, ARM = \angle \, ARC = 180^{\circ} - \angle \, ABC = \angle \, QBC = \angle \, QBP = \beta$$
In triangle $\Delta \, BQP$
$$\angle \, BQP = 180^{\circ} - \angle \, BPQ - \angle \, QBP = 180^{\circ} - \alpha - \beta$$
Thus in quad $AQPR$ $$\angle \, PRA + \angle \, AQP = (\angle \, PRM + \angle \, ARM) + \angle \, BQP = (\alpha + \beta) + (180^{\circ} - \alpha - \beta) = 180^{\circ}$$
The quad $AQPR$ is inscribed in a circle.
Because of the perpendicularity between $c_P$ and $\omega$, when inverted in circle $c_P$, point $A$ is mapped to point $D$, point $C$ is mapped to point $B$ and point $R$ is mapped to point $L$. By Lemma 2, circle $c(CDPG)$ is mapped to the line $AB$ and since $G$ is from $c(CDPG) \, \cap \, PQ$ and $Q$ is on $AB$, point $G$ is inverted into point $Q$. By inversion in $c_P$ and Lemma 2, since the circumcircle $c(AQPR)$ of quad $AQPR$ passes through the center $P$ of $c_P$, it is inverted into a straight line that contains the inverse images of points $R, \, A, \, Q$, which happen to be $L, \, D, \, G$. Therefore the points $L, \, D, \, G$ are collinear.
Remark (outside the scope of the problem): The segment $OG$ is perpendicular to $PQ$ because: first of all, the point $O$ lies on the radical axis $E_1E_2$ of the three circles $c_P, c_Q$ and $c_M$ due to the fact that $\omega$ is perpendicular to all three of them. Then, after inversion in $c_P$, the line $PQ$ is mapped to itself and the circle $c_M$ is mapped to the straight line $E_1E_2$ (Lemma 1) and $E_1E_2$ is perpendicular to $PQ$ (because of Lemma 1 or also because $c_M \, \perp \, PQ$ and so should be its image $E_1E_2$). However, we have already established that $Q$ is inverted into $G$, i.e. $G$ also lies on $E_1E_2$ so $E_1E_2$ intersects $PQ$ in $G$ and therefore $OG \equiv E_1E_2$ is perpendicular to $PQ$.
