On the plane there are two non-intersecting circles with equal radii and with centres $O_1$ and $O_2$, line $s$ going through these centres, and their common tangent $t$. The third circle is tangent to these two circles in points $K$ and $L$ respectively, line $s$ in point $M$ and line $t$ in point $P$. The point of tangency of line $t$ and the first circle is $N$.
a) Find the length of the segment $O_1O_2$.
b) Prove that the points $M, K$ and $N$ lie on the same line
### Part (a): Find the length of the segment \( O_1O_2 \)

1. **Given Data:**
   - Two non-intersecting circles with equal radii \( r \) and centers \( O_1 \) and \( O_2 \).
   - The line \( s \) passes through the centers \( O_1 \) and \( O_2 \).
   - The common tangent \( t \) touches both circles.

2. **Positioning the Circles:**
   - Place the first circle with center \( O_1 \) at the origin, i.e., \( O_1(0,0) \).
   - The second circle has the same radius \( r \) and is positioned such that it does not intersect the first circle. Let the center of the second circle be \( O_2 \).

3. **Distance Between Centers:**
   - Since the circles are non-intersecting and have equal radii, the distance between the centers \( O_1 \) and \( O_2 \) must be at least \( 2r \).
   - The exact distance is \( 2r \) because the circles are tangent to the third circle, which is tangent to both circles at points \( K \) and \( L \).

Thus, the length of the segment \( O_1O_2 \) is:
\[ O_1O_2 = 2r \]

### Part (b): Prove that the points \( M, K \), and \( N \) lie on the same line

1. **Given Data:**
   - The third circle is tangent to the two given circles at points \( K \) and \( L \).
   - The third circle is also tangent to the line \( s \) at point \( M \) and to the common tangent \( t \) at point \( P \).
   - The point of tangency of line \( t \) and the first circle is \( N \).

2. **Positioning the Third Circle:**
   - Let the center of the third circle be \( O_3 \) and its radius be \( \frac{r}{2} \).
   - Since the third circle is tangent to the line \( s \) at \( M \), \( M \) is directly below \( O_3 \) on the line \( s \).

3. **Coordinates of Points:**
   - The center \( O_3 \) of the third circle is at \( (\sqrt{2}r, \frac{r}{2}) \).
   - The point \( M \) is directly below \( O_3 \) on the line \( s \), so \( M \) has coordinates \( (\sqrt{2}r, 0) \).
   - The point \( N \) is the point of tangency of line \( t \) with the first circle, so \( N \) has coordinates \( (0, r) \).

4. **Equation of Line \( MN \):**
   - The slope of the line \( MN \) is:
     \[
     \text{slope of } MN = \frac{r - 0}{0 - \sqrt{2}r} = -\frac{1}{\sqrt{2}}
     \]
   - The equation of the line \( MN \) passing through \( M(\sqrt{2}r, 0) \) is:
     \[
     y - 0 = -\frac{1}{\sqrt{2}}(x - \sqrt{2}r)
     \]
     Simplifying, we get:
     \[
     y = -\frac{1}{\sqrt{2}}x + r
     \]

5. **Point \( K \) on Line \( MN \):**
   - The point \( K \) lies on the third circle and on the line \( O_1O_3 \).
   - The coordinates of \( K \) are \( \left(\frac{2\sqrt{2}}{3}r, \frac{r}{3}\right) \).
   - Substitute \( K \) into the equation of line \( MN \):
     \[
     \frac{r}{3} = -\frac{1}{\sqrt{2}}\left(\frac{2\sqrt{2}}{3}r\right) + r
     \]
     Simplifying, we get:
     \[
     \frac{r}{3} = -\frac{2r}{3} + r
     \]
     \[
     \frac{r}{3} = \frac{r}{3}
     \]
   - This confirms that \( K \) lies on the line \( MN \).

Thus, the points \( M, K \), and \( N \) lie on the same line.

\(\blacksquare\)

The final answer is \( \boxed{ O_1O_2 = 2r } \) and the points \( M, K \), and \( N \) lie on the same line.