Let's look at it somewhat backwards. If $X$ is fixed, there is clearly only one circle satisfying the conditions, so let's show that if $BX$ bisects the chord $MN$ of the circumcircle of $ABC$, then $M,N$ are the points we are looking for. $BNXM$ is a parallelogram, so the entire figure is symmetric wrt the midpoint of $BX$. Since the parallel through $B$ to $AC$ is tangent to $(ABC)$, it means that $AC$ is tangent to $(MXN)$, so $M,N$ are the points mentioned in the problem. We have shown that, if $T$ is the midpt of $BX$, then $P,Q$ are the intersections of $BA,BC$ with the perpendicular through $T$ to $OT$ (call this property $(*)$). On the other hand, if we take $P'\in BA,Q'\in BC$ s.t. $XP'\|BC,XQ'\|BA$, then it's easy to see that they satisfy $(*)$ and $\angle P'OQ'=\pi-\angle ABC$, so $P=P',Q=Q'$ and $\angle POQ=\pi-\angle ABC$, which is what we wanted.

Can you please explain to me why : "If $ X $ is fixed, there is clearly only one circle satisfying the conditions " ?

I have just solved a generalized version of this problem in http://www.mathlinks.ro/Forum/viewtopic.php?t=20669 post #4. In fact, in that post, I proved the following: Construct the circle passing through the vertices A and B of triangle ABC and touching the side BC at the point B, and construct the circle passing through the vertices B and C and touching the side AB at the point B. These two circles intersect at B; let their second point of intersection be called F. Then, the circumcircle of triangle PBQ always passes through the point F. This actually holds for arbitrary triangles ABC. Now, for isosceles triangles ABC such that AB = BC, the point F can be characterized in a simpler way: It is the circumcenter of triangle ABC. In order to prove this, note that, at first, the point F was defined in a way symmetric with respect to the vertices C and A; thus, since the triangle ABC is isosceles with base CA, the point F must lie on the symmetry axis of this isosceles triangle, i. e. on the perpendicular bisector of the segment CA. Hence, FC = FA. Now, as I showed in the above-mentioned post, the triangles FAB and FBC are directly similar; hence, FA : FB = FB : FC, so that $FA\cdot FC=FB^2$. Since FC = FA, this becomes $FA\cdot FA=FB^2$, so that $FA^2=FB^2$ and thus FA = FB. Combining this with FC = FA, we obtain FA = FB = FC, and thus the point F is the circumcenter of triangle ABC. So it follows that the circumcircle of triangle PBQ always passes through the circumcenter of triangle ABC. And the problem is solved. nttu wrote: Can you please explain to me why : "If $ X $ is fixed, there is clearly only one circle satisfying the conditions " ? Well, I don't understand why Grobber finds this so obvious, but it follows from one of my observations in http://www.mathlinks.ro/Forum/viewtopic.php?t=20669 post #4, actually from the following one: darij grinberg wrote: First, it's time to involve the assumption that the radical axis of the circle k [note: this is our circle through the point X touching the side AC] and the circumcircle of triangle ABC bisects the segment BX. In other words, if M is the midpoint of the segment BX, then this point M has equal powers with respect to the circle k and to the circumcircle of triangle ABC. Again, let's compute these powers: If the line BX meets the circle k at a point U (apart from X) and the circumcircle of triangle ABC at a point T (apart from B), then the power of the point M with respect to the circle k is $MX\cdot MU$, and the power of the point M with respect to the circumcircle of triangle ABC is $MB\cdot MT$. Since the two powers of M are equal, we thus have $MX\cdot MU=MB\cdot MT$. In other words, $MX\cdot MU=BM\cdot TM$. But since M is the midpoint of the segment BX, we have MX = BM, and thus MU = TM. Hence, BU = BM + MU = MX + TM = TX, In fact, this yields an easy construction of the circle k: Construct the point T as the point where the line BX meets the circumcircle of triangle ABC (apart from B), and construct the point U as the point on the line BX such that BU = TX (there is exactly one such point U, since we are working with directed segments). Then, the circle k must pass through the points X and U and touch the line AC at the point X; this is already enough for uniquely determining the circle k (you can construct its center as the point where the perpendicular to AC at X meets the perpendicular bisector of the segment XU). So the circle k is unique. Darij