I am doubting how to prove this problem:
Two circles intersect in the points $A$ and $B .$ A variable line through $A$ intersects the circles again in $P$ and $Q .$ If $R$ divides the segment $P Q$ in a given ratio, show that the locus of $R$ is a circle.
I did a construction in geogebra:
Should I just say that is it proved immediately from the circle of Apollonius theorem? It states:
Thanks in advance.
Without loss of generality, we can assume that $O = (0,0)$, so that the equation of the first circle is $ p^T p = R_1^2 $, while the equation of the second circle is $(p - O')^T (p - O') = R_2^2 $. Let the parametric equation of the variable line be
$ q = A + s u $ , where $u = [\cos \theta, \sin \theta]^T$, and $s \in \mathbb{R} $
Now we'll find points $P$ and $Q$. $P$ is the intersection of the line with the first circle, so
$$(A + s u)^T (A + s u) = R_1^2 $$
since $A^T A = R_1^2$, then
$ 2 s u^T A + s^2 = 0$
Therefore,
$ s = - 2 u^T A$
Thus, $ P = A - 2 (u^T A) u$
similarly for the second circle, we have,
$$ ( A - O' + s u )^T ( A - O' + s u ) = R_2^2$$
and since $(A - O')^T (A - O') = R_2^2 $ , then
$ s = - 2 u^T (A - O')$
Thus,
$Q = A - 2 u^T (A - O') u$
Assuming the fixed ratio $r = \dfrac{PR}{PQ}$, then
$\begin{equation} \begin{split} R &= P + r (Q - P) \\ &= A - 2 (u^T A) u + r ( - 2 ) u^T ( A - O' - A ) u \\ &= A - 2 (u^T A ) u + 2 r u^T (O') u \\ &= A + 2 u^T (r O' - A) u \\ \end{split} \end{equation}$
Now assume that $r O' - A = v = [v_1, v_2]^T $, then
$\begin{equation} \begin{split} R &= A + 2 ( v_1 \cos \theta + v_2 \sin \theta ) [ \cos \theta, \sin \theta]^T \\ &= A + 2 [ v_1 \cos^2 \theta + v_2 \sin \theta \cos \theta , v_1 \cos \theta \sin \theta + v_2 \sin^2 \theta ]^T \\ &= A + [ v_1 (1 + \cos 2 \theta) + v_2 \sin 2 \theta , v_1 \sin 2 \theta + v_2 (1 - \cos 2 \theta ) ^T \\ &= A + [v_1, v_2]^T + \cos 2 \theta [v_1, -v_2]^T + \sin 2 \theta [ v_2, v_1 ]^T \end{split} \end{equation} $
Since the vectors $[v_1, -v_2]^T$ and $[v_2, v_1]^T$ are orthogonal and of equal length, then this is clearly the equation of a circle with center $A + [v_1 , v_2]^T $ and radius $R = \sqrt{v_1^2 + v_2^2 } = \| r O' - A \| $