Question:

**Step 1: Identify the line formed by the midpoints** The common tangents $PQ$ and $ST$ touch $C_1$ and $C_2$. The length of a tangent segment from a point on the line joining the midpoints to $C_1$ is equal to the length of the tangent segment to $C_2$ (since the midpoint of $PQ$ has equal tangents to $C_1$ and $C_2$). Thus, the line joining the midpoints of the common tangents is the radical axis of the two circles. **Step 2: Find the equation of the radical axis** The equation of $C_1$ is $x^2 + y^2 - 1 = 0$.\nThe equation of $C_2$ is $(x-4)^2 + (y-1)^2 - r^2 = 0 \implies x^2 + y^2 - 8x - 2y + 17 - r^2 = 0$.\nThe radical axis is given by $C_1 - C_2 = 0$, which is $8x + 2y - 18 + r^2 = 0$. **Step 3: Find the coordinates of point B** The radical axis meets the $x$-axis at point $B$. Setting $y=0$ in the radical axis equation gives $8x - 18 + r^2 = 0 \implies x = \frac{18 - r^2}{8}$.\nSo, $B = \left( \frac{18 - r^2}{8}, 0 \right)$. **Step 4: Calculate the value of r^2** We are given that $AB = \sqrt{5}$, where $A = (4, 1)$.\nUsing the distance formula: $AB^2 = \left( \frac{18 - r^2}{8} - 4 \right)^2 + (0 - 1)^2 = 5$.\n$\left( \frac{18 - r^2 - 32}{8} \right)^2 + 1 = 5 \implies \left( \frac{-14 - r^2}{8} \right)^2 = 4$.\nTaking the square root: $\frac{-14 - r^2}{8} = \pm 2$.\nCase 1: $-14 - r^2 = 16 \implies r^2 = -30$ (Not possible).\nCase 2: $-14 - r^2 = -16 \implies r^2 = 2$.\nSince $1 < r < 3 \implies 1 < r^2 < 9$, the value $r^2 = 2$ is valid.