Two circles $ C_1,C_2$ with different radii are given in the plane, they touch each other externally at $ T$. Consider any points $ A\in C_1$ and $ B\in C_2$, both different from $ T$, such that $ \angle ATB \equal{} 90^{\circ}$.

(a) Show that all such lines $ AB$ are concurrent.
(b) Find the locus of midpoints of all such segments $ AB$.
### Part (a)

1. **Define the centers and coordinates:**
   Let \( O_1 \) and \( O_2 \) be the centers of circles \( C_1 \) and \( C_2 \) respectively. Assume without loss of generality that \( T = (0,0) \), \( O_1 = (-y, 0) \), and \( O_2 = (x, 0) \), where \( y \) and \( x \) are positive real numbers representing the radii of the circles.

2. **Equations of the circles:**
   The equations of the circles are:
   \[
   C_1: (a + y)^2 + b^2 = y^2
   \]
   \[
   C_2: (a - x)^2 + b^2 = x^2
   \]

3. **Points on the circles:**
   Let \( A = (m, n) \in C_1 \) and \( B = (p, q) \in C_2 \).

4. **Lines through \( T \):**
   The lines \( AT \) and \( BT \) are:
   \[
   l_1: y = \frac{n}{m} x
   \]
   \[
   l_2: y = -\frac{m}{n} x
   \]
   since \( \angle ATB = 90^\circ \).

5. **Point \( B \) on \( l_2 \):**
   Since \( B \in l_2 \), we have:
   \[
   q = -\frac{m}{n} p
   \]

6. **Substitute \( B \) into \( C_2 \):**
   Substitute \( B = (p, -\frac{m}{n} p) \) into the equation of \( C_2 \):
   \[
   (p - x)^2 + \left( -\frac{m}{n} p \right)^2 = x^2
   \]
   Simplifying, we get:
   \[
   (p - x)^2 + \frac{m^2}{n^2} p^2 = x^2
   \]
   \[
   p^2 \left( 1 + \frac{m^2}{n^2} \right) - 2px + x^2 = x^2
   \]
   \[
   p^2 \left( 1 + \frac{m^2}{n^2} \right) = 2px
   \]
   \[
   p = \frac{2xn^2}{m^2 + n^2}
   \]

7. **Equation of line \( AB \):**
   The line \( AB \) can be written as:
   \[
   \overline{AB} = l_3: y = -\frac{2xmn + m^2n + n^3}{2xn^2 - m^3 - mn^2} x + \frac{2xn(n^2 + m^2)}{2xn^2 - m^3 - mn^2}
   \]

8. **Intersection with the \( x \)-axis:**
   The intersection of \( l_3 \) with the \( x \)-axis occurs when \( y = 0 \):
   \[
   x' = \frac{2x(n^2 + m^2)}{2xm + m^2 + n^2} = -2 \frac{xy}{x - y}
   \]
   This shows that the intersection point depends only on the radii \( x \) and \( y \), not on the specific points \( A \) and \( B \). Therefore, all such lines \( AB \) are concurrent.

### Part (b)

1. **Circumcenter of \( \triangle ATB \):**
   Let \( M \) be the midpoint of \( AB \). Since \( \angle ATB = 90^\circ \), \( M \) is the circumcenter of \( \triangle ATB \).

2. **Perpendicular bisectors:**
   \( O_1M \) is the perpendicular bisector of \( AT \) and \( O_2M \) is the perpendicular bisector of \( BT \).

3. **Angle between bisectors:**
   Since \( \angle O_1MO_2 = 90^\circ \), \( M \) lies on the circle with diameter \( O_1O_2 \).

4. **Locus of midpoints:**
   The locus of the midpoints \( M \) of all such segments \( AB \) is a circle with diameter \( O_1O_2 \).

The final answer is the locus of midpoints is a circle with diameter \( \boxed{ O_1O_2 } \).