Step-by-step Derivations

1. Identify the given information:

   • Circle $C_1$: Center $O = (0,0)$, radius $r_1 = 1$. Its equation is $x^2 + y^2 = 1$.
   • Circle $C_2$: Center $A = (4,1)$, radius $r_2 = r$. Its equation is $(x-4)^2 + (y-1)^2 = r^2$.
   • The range for the radius $r$ is $1 < r < 3$.
   • $PQ$ and $ST$ are two distinct common tangents. $P, S$ are on $C_1$ and $Q, T$ are on $C_2$.
   • A line is formed by joining the midpoints of $PQ$ and $ST$. Let's call these midpoints $M$ and $N$ respectively.
   • This line $MN$ intersects the x-axis at a point $B$.
   • The distance between points $A$ and $B$ is given as $AB = \sqrt{5}$.

2. Find the locus of the midpoint of a common tangent segment: Let's consider one of the common tangents, say $PQ$. Let $M(x,y)$ be its midpoint.

   • $P$ is the point of tangency on $C_1$, so the radius $OP$ is perpendicular to the tangent line $PQ$. Thus, $\triangle OMP$ is a right-angled triangle with the right angle at $P$.
   • $Q$ is the point of tangency on $C_2$, so the radius $AQ$ is perpendicular to the tangent line $PQ$. Thus, $\triangle AMQ$ is a right-angled triangle with the right angle at $Q$.

   Using the Pythagorean theorem in these two triangles:

   • In $\triangle OMP$: $OM^2 = OP^2 + PM^2$. Since $OP = r_1 = 1$, we have $OM^2 = 1^2 + PM^2 = 1 + PM^2$.
   • In $\triangle AMQ$: $AM^2 = AQ^2 + QM^2$. Since $AQ = r_2 = r$, we have $AM^2 = r^2 + QM^2$.

   Since $M$ is the midpoint of the segment $PQ$, we have $PM = QM$, which implies $PM^2 = QM^2$. From the two equations above, we can write: $PM^2 = OM^2 - 1$ $QM^2 = AM^2 - r^2$ Equating these two expressions for $PM^2$ and $QM^2$ gives: $OM^2 - 1 = AM^2 - r^2$ $OM^2 - AM^2 = 1 - r^2$

   This equation describes the locus of the midpoint $M$. This relation holds true for the midpoints of both direct and transverse common tangents.

3. Determine the equation of the line: Let the coordinates of the midpoint $M$ be $(x,y)$. The coordinates of the centers are $O(0,0)$ and $A(4,1)$. $OM^2 = (x-0)^2 + (y-0)^2 = x^2 + y^2$ $AM^2 = (x-4)^2 + (y-1)^2 = x^2 - 8x + 16 + y^2 - 2y + 1 = x^2 + y^2 - 8x - 2y + 17$

   Substituting these into the locus equation $OM^2 - AM^2 = 1 - r^2$: $(x^2 + y^2) - (x^2 + y^2 - 8x - 2y + 17) = 1 - r^2$ $8x + 2y - 17 = 1 - r^2$ $8x + 2y = 18 - r^2$ $4x + y = \frac{18 - r^2}{2}$

   This is the equation of the line containing the midpoints of all common tangent segments. The line joining the midpoints of $PQ$ and $ST$ is this very line.

4. Find the coordinates of point B: The point $B$ is the intersection of this line with the x-axis. The equation of the x-axis is $y=0$. Substitute $y=0$ into the line equation: $4x_B + 0 = \frac{18 - r^2}{2}$ $x_B = \frac{18 - r^2}{8}$ So, the coordinates of point $B$ are $\left( \frac{18 - r^2}{8}, 0 \right)$.

5. Use the given distance to find r²: We are given that the distance $AB = \sqrt{5}$. The coordinates of $A$ are $(4,1)$. $AB^2 = (x_B - x_A)^2 + (y_B - y_A)^2 = (\sqrt{5})^2 = 5$. $\left( \frac{18 - r^2}{8} - 4 \right)^2 + (0 - 1)^2 = 5$ $\left( \frac{18 - r^2 - 32}{8} \right)^2 + 1 = 5$ $\left( \frac{-r^2 - 14}{8} \right)^2 = 4$ $\left( \frac{-(r^2 + 14)}{8} \right)^2 = 4$ $\frac{(r^2 + 14)^2}{64} = 4$ $(r^2 + 14)^2 = 256$

   Taking the square root of both sides: $r^2 + 14 = \pm \sqrt{256}$ $r^2 + 14 = \pm 16$

   This gives two possibilities: Case 1: $r^2 + 14 = 16 \implies r^2 = 2$. Case 2: $r^2 + 14 = -16 \implies r^2 = -30$. This is not possible since $r^2$ must be positive.

   So, the only valid solution is $r^2 = 2$.

6. Verify the condition on r: The problem states that $1 < r < 3$. If $r^2 = 2$, then $r = \sqrt{2}$. Since $1 < \sqrt{2} \approx 1.414 < 3$, the condition is satisfied.

Thus, the value of $r^2$ is 2.