Let $l'$ be the line through $P \perp$ to $l$. Let $l' \cap \Gamma = \{A,B\}$. Let $K$ and $L$ be the 2 midpoints of arc $AB$ of $\Gamma$. Let $KP \cap \Gamma =\{K,X\}$ and let $LP \cap \Gamma = \{L,Y\}$ Let $OX\cap l = M$ and $OY\cap l = N$. Then consider the 2 circles- $(M,MP)$ and $(N,NP)$. These are the 2 required circles.

Draw line through center O of tau parallel to l and it meets tau at A and B. Then let AP, BP meets tau at X, Y and OX, OY meets l at K, L. Then circles are with center K, L...

Invert the figure in $P$ and then construct the tangent to $\tau$ which is perpendicular to $l$. Let this tangent meet $l$ at $A$ . Then the required circle is one with diameter as $P$ and the inverse of $A$. The proof directly follows from properties of inversion.

Let,$G$ denote the map inversion of the circle center at a point$X$. For Any $X$ ,$G(X)$ is drawable with rular and compass. So,we can draw $G(\tau)$. Now,Let $\Gamma$ be our needed circle. So,$G(\Gamma)$ is a line perpendicular to the line $l=G(l)$.which touch the circle$\tau$. So,we can draw two perpendicular line on a fixed point of a straight line from another external line . So,these line are images with mapping $G$. We can find their pre images . So ,two such configuration is possible.