Si una recta trazada por un punto de intersección de dos circunferencias corta nuevamente a las circunferencias en $P$ y $Q$, respectivamente, desmuéstrese que las circunferencias con centros P y Q, cada una ortogonal a la otra, son ortogonales entre sí.
Attempt traslation:
A line drawn through the point of intersection of two circles $C_1$ and $C_2$ intersects the circles again at $P$ and $Q$ respectively. Show that if the circles with centers $P$ and $Q$ are orthogonal to $C_2$ and $C_1$, respectively, they are also orthogonal to each other.
My progress is at follows:
In the figure, let
- $C_1$ be the blue circumference and $O_1$ its center.
- $C_2$ be the green circumference and $O_2$ its center.
Let $E$ be their meet point. $PQ$ is an aribitrary line through $E$, as in the problem.
Now,
- $D_1$, red circumference, with center in $P$, is orthogonal to $C_2$ and $C$ is their intersection point
- $D_2$, pink circumference, with center in $Q$, is orthogonal to $C_1$ and $G$ is their intersection point
Then, $\angle O_{1}GQ = \angle O_{2}CP = 90^{\circ}$
Let $H$ be the intersection point of $D_1$ and $D_2$, I want to show that $\angle PHQ = 90^{\circ}$, wich is the same, $\left(\overline{PQ}\right)^{2} = \left(\overline{PH}\right)^{2} + \left(\overline{HQ}\right)^{2}$
$D_{1}\bot C_{2} $ implies that $$\left(\overline{O_{2}P}\right)^{2} = \left(\overline{PC}\right)^{2} + \left(\overline{CO_{2}}\right)^{2} = \left(\overline{PH}\right)^{2} + \left(\overline{O_{2}Q}\right)^{2}$$
Analogously,
$D_{2}\bot C_{1} $ implies that $$\left(\overline{O_{1}Q}\right)^{2} = \left(\overline{QG}\right)^{2} + \left(\overline{GO_{1}}\right)^{2} = \left(\overline{HQ}\right)^{2} + \left(\overline{O_{1}P}\right)^{2}$$
By adding and rearranging
$$\left(\overline{PH}\right)^{2} + \left(\overline{HQ}\right)^{2} =\left(\overline{O_{2}P}\right)^{2} + \left(\overline{O_{1}Q}\right)^{2} - \left(\overline{O_{2}Q}\right)^{2} -\left(\overline{O_{1}P}\right)^{2} $$
How can I continue? It's the right way?
P.S.I apologize for possible misspellings, indeed I don´t speak english
$\def\C{{\cal C}}$ Let $r_1,r_2,r_P,r_Q$ be the radii of the circles ${\C_1,\C_2,\C_P,\C_Q}$ centered at $O_1,O_2,P,Q$, respectively.
From the orthogonality of $\C_P$ and $\C_2$, and that of $\C_Q$ and $\C_1$ it follows: $$ PO_2^2=r_P^2+r_2^2,\quad QO_1^2=r_Q^2+r_1^2.\tag1 $$
Now by the power of a point theorem we have: $$ \begin{align} PE\cdot PQ&=PO_2^2-r_2^2=r_P^2\tag{2a}\\ QE\cdot QP&=QO_1^2-r_1^2=r_Q^2\tag{2b}\\ \end{align} $$ Summing the two equalities one obtains: $$ PQ^2=r_P^2+r_Q^2,\tag3 $$ which is equivalent to the claim.