Take the point $X$ on $AB$ s.t. $(B,C;A,X)=-1$. Now let $U=BM\cap CN,\ V=BN\cap CM$. The line $UV$ passes through $X$ (because of well-known projective properties of the complete quadrilateral). Let $S=UV\cap MN$. We have $(BM,BN;BA,BS)=-1$, and since $A$ is the midpoint of $MN$, we find $BS\|MN\Rightarrow UV\|MN$, so $X$ is the midpoint of $UV$. Furthermore, we have $\frac{XU}{AN}=\frac{CX}{CA}$, which is fixed, so $XU$ is constant, meaning that $U$ lies on a circle obtained from $T$ by a homothety of centers $B$ and $C$ (one of them is the external center, the other one is the internal center).

I assume that the points A, B, C and the radius r are fixed, while the two diametrically opposite points M and N move along the circle T. If this is right, then it's a quite nice problem. In fact, I will only consider the case when the point B lies between the points A and C (in all other cases, the solution is analogous). Let D be the harmonic conjugate of the point A with respect to the segment BC. Then the points B and C are harmonic conjugates of each other with respect to the segment AD. Thus, $\frac{AC}{DC}=-\frac{AB}{DB}$, with directed segments. Let p be a number such that $\frac{r}{p}=\frac{AC}{DC}=-\frac{AB}{DB}$, and consider the circle P(D, p). Now I claim that this circle P is the required locus of the point of intersection of the lines BM and CN. In order to prove this, I denote the point of intersection of the lines BM and CN by K and aim at proving that this point K lies on the circle P. In fact, since the segment MN is a diameter of the circle T, it passes through the point A, and we have MA = AN. Now, let the parallel to this diameter MN through the point D meet the line CN at U. Then, by Thales, $\frac{AN}{DU}=\frac{AC}{DC}$. On the other hand, let the parallel to the diamater MN through the point D meet the line BM at V. Then, Thales yields $\frac{AM}{DV}=\frac{AB}{DB}$. Since $\frac{AC}{DC}=-\frac{AB}{DB}$, we have $\frac{AN}{DU}=-\frac{AM}{DV}$, or, equivalently, $\frac{AN}{DU}=\frac{MA}{DV}$. Since MA = AN, this implies DU = DV. This implies that the points U and V coincide, i. e. the parallel to the diameter MN through the point D meets the lines CN and BM at the same point. This point, of course, must then coincide with the point of intersection K of the lines CN and BM. Thus, K = U = V. Therefore, the equation $\frac{AN}{DU}=\frac{AC}{DC}$ becomes $\frac{AN}{DK}=\frac{AC}{DC}$. On the other hand, we know that $\frac{r}{p}=\frac{AC}{DC}$; thus, $\frac{AN}{DK}=\frac{r}{p}$. Now, AN = r; hence, DK = p, and it follows that the point K lies on the circle P(D, p), completing our proof. [Note that from $\frac{r}{p}=\frac{AC}{DC}=-\frac{AB}{DB}$, it follows that the points B and C are the two centers of similitude of the circles T and P.] Darij

You have good solutions Grobber and Darij. All students who solve it used the idea with harmonic division. But in fact the solution given by the propozers was bazed on analitic geometry and it is long and not so beautiful.