IMO Hk TST, 1997

a.
Let $O_1$ and $O_2$ be the centres of $C_1$ and $C_2$ respectively. Let $O_1O_2$ meet $C_1$ and $C_2$ again at $C$ and $D$ respectively. Note that $CT$ and $DT$ are diameters of the two circles. Therefore, we have
$$\angle CAT = \angle BTA = \angle TBD = 90^\circ.$$
This implies $AC \parallel BT$ and $AT \parallel BD$, and so $\triangle ACT \sim \triangle BTD$. Since the radii are distinct, the triangles are not congruent. Thus, $AB$ meets $CD$ at a point $X$. Note that $\triangle XAC \sim \triangle XBT$. Using the similar triangles, we obtain
$$\frac{XT}{XC} = \frac{BT}{AC} = \frac{TD}{CT},$$
which is a fixed ratio. Therefore, $X$ is a fixed point (as directed lengths are used). In other words, all such lines $AB$ are concurrent at $X$.

b.
The locus is the circle with diameter $O_1O_2$.
Let $M$ be the midpoint of $AB$. Since $M$ and $O_1$ are the midpoints of $AB$ and $CT$, we have $AC \parallel MO_1 \parallel BT$. Similarly, we have $AT \parallel MO_2 \parallel BD$. This implies $\angle O_1MO_2 = 90^\circ$, and hence $M$ lies on the circle $\Gamma$ with diameter $O_1O_2$.
When $A$ approaches $C$ from above, the point $B$ approaches $T$ from above. This shows $M$ approaches $O_1$ from above. When $A$ approaches $C$ from below, the point $B$ approaches $T$ from below. This shows $M$ approaches $O_1$ from below. Also, $M = O_1$ only when $A = C$ and $B = T$. By continuity, $M$ runs through all points on $\Gamma$, and so $\Gamma$ is the locus of $M$.