Two circles $c$ and $d$ that intersect at points $A$ and $B$ are given. Let $p$ be a line passing through $A$ that intersects circles $c$ and $d$ at points $P_1$ and $P_2$. Construct a circle tangential to $p$ and orthogonal to both $c$ and $d$.
What I did was that I constructed a circle $k=k(P, |PS_1|)$ so that $P$ is the midpoint of $S_1S_2$. Then I applied inversion about $k$, which maps circles $c$ and $d$ to two parallel lines $c'$ and $d'$, and line $p$ to circle $p'$ that passes through point $P$. Then I found line $l$ such that it's perpendicular to $c'$ and $d'$ and tangential to circle $p '$. Finally, I applied inversion about $k$ again so it would map $l$ to circle $l'$. I thought $l'$ would be the solution, but apparently it isn't. Why doesn't this work and what should I do instead?
EDIT: Now I see what a dumb mistake I made, so I'm sorry for hurting yalls eyes. I still wonder how this problem could be solved using inversion. I was thinking to apply inversion about circle centered at $A$ that would map both $c$ and $d$ to lines $c'$ and $d'$ and it would map line $p$ to itself, but I don't see how that would be helpful
Best Answer
In fact, after having presented an analytical solution, I have discovered a "pure geometry" simple construction using inversion.
Let us take the notations of figure below $O_1,O_2$ for the centers of circles $(C_1),(C_2)$ resp. with intersection points $A,A'$, $AD_1,AD_2$ for diameters.
Let us proceed by necessary condition first. Let $F$ be the center of orthogonal circle $(C)$ to both circles $(C_1),(C_2)$ with radius $FE \perp (p)$. Let $(I)$ be the inversion with inversion circle $(C)$. The two circles $(C_1),(C_2)$, being orthogonal to $(C)$ are globally invariant by $(I)$. Therefore $A,A'$ are exchanged by $(I)$, which means two properties : $A,A',F$ are aligned and
$$FA.FA'=FE^2\tag{1}$$
But (1) is possible iff $EB$ is the altitude issued from $E$ in the right triangle $AEF$ (see second "mean proportional" property here) which is a characteristic property of the main altitude of a right triangle.
Whence the construction of point $F$ :
Draw the perpendicular line in $A'$ to $AA'$ (aka radical axis of circles $(C_1),(C_2)$). This line intersects line (p) in $E$. Then erect in $E$ the perpendicular line to (p). This line intersects line $AA'$ in $F$, and it's finished.
Once $F$ is known, the radius of the orthogonal circle is of course $FE$.
Remark : I still don't understand the construction of the other answer.