Let $O_{1},\ O_{2}$ be centers and $r_{1},\ r_{2}$ radii of $\Gamma_{1},\ \Gamma_{2}.$ Tangent r of $\Gamma_{1}$ at M is the radical axis of $\Gamma_{1},\ \Gamma_{2},$ it cuts BP in half at $N \in l$, BP = 2BN. NB = NP = NM, hence $BM \perp PM,$ but AB is a diameter of $\Gamma_{1}$, $BM \perp AM$ and A, M, P are collinear, AM cuts $l$ at P. Let $d \parallel l$ cut AB at D, different from $O_{1}$, such that $BD = BO_{1}$ and let $O_{2}P \perp l$ cut d at Q. $O_{1}O_{2}= r_{1}+r_{2}= QO_{2},$ $O_{2}$ lies on a parabola $\mathcal P$ with focus $O_{1},$ vertex B and directrix d. Inversion in a circle (A) with radius AB carries $\Gamma_{1}$ to $l,$ $l$ to $\Gamma_{1}$ and $\Gamma_{2}$ to a circle tangent to $\Gamma_{1},\ l$ at M, P, i.e., into itself, hence $\Gamma_{2}\perp (A)$ for any M.

Two circles are orthogonal iff the triangle formed by their centers $ O_{3}$, $ O_{2}$, and either of their points of intersection $ T$ is rectangle at $ T$, or iff $ O_{3}O_{2}^{2}= r_{3}^{2}+r_{2}^{2}$, where $ O_{3}$, $ O_{2}$ are the centers of the circles and $ r_{3}$, $ r_{2}$ their respective radii. We just need to show that taking $ O_{3}= A$, $ r_{3}= 2r_{1}$, then $ AO_{2}^{2}= r_{2}^{2}+4r_{1}^{2}$, or since $ AO_{2}^{2}= BP^{2}+(r_{2}-2r_{1})^{2}$, we just need to prove that $ BP^{2}= 4r_{1}r_{2}$. But this is easy, since $ (r_{1}+r_{2})^{2}= O_{1}O_{2}^{2}= BP^{2}+(r_{2}-r_{1})^{2}$, and we are done.