Let $C_1$ be the circle of radius 1 with center at the origin. Let $C_2$ be the circle of radius $r$ with center at the point $A=(4,1)$, where $1<r<3$. Two distinct common tangents $PQ$ and $ST$ of $C_1$ and $C_2$ are drawn. The tangent $PQ$ touches $C_1$ at $P$ and $C_2$ at $Q$. The tangent $ST$ touches $C_1$ at $S$ and $C_2$ at $T$. Mid points of the line segments $PQ$ and $ST$ are joined to form a line which meets the $x$-axis at a point $B$. If $AB = \sqrt{5}$, then the value of $r^2$ is

Let C₁ be the circle of radius 1 with center at the origin. Let C₂ be the circle of radius r with center at the point A=(4,1), where 1<r<3. Two distinct common tangents PQ and ST of C₁ and C₂ are drawn. The tangent PQ touches C₁ at P and C₂ at Q. The tangent ST touches C₁ at S and C₂ at T. Mid points of the line segments PQ and ST are joined to form a line which meets the x-axis at a point B. If AB = √{5}, then the value of r² is $\text{Given: } C_1:\ \text{center } O(0,0),\ \text{radius } 1$ $C_2:\ \text{center } A(4,1),\ \text{radius } r,\quad 1 < r < 3$ $\text{common tangents } PQ,\ ST \Rightarrow \text{midpoints joined meet } x\text{-axis at } B$ $\text{To find: } r^2,\ \text{given } AB = \sqrt{5}$ $1.\ \text{draw the circles and their common tangents}$ $2.\ \text{tangent } x\cos\theta + y\sin\theta = 1$ $3.\ \text{line of midpoints, then } B, \text{ then } r^2$ $C_1:\ x^2 + y^2 = 1$ $C_2:\ (x-4)^2 + (y-1)^2 = r^2$ $OA = \sqrt{4^2 + 1^2} = \sqrt{17}$ $\text{two direct common tangents } PQ,\ ST$ $\text{tangent to } C_1:\ x\cos\theta + y\sin\theta = 1$ $\text{point of contact } P = (\cos\theta,\ \sin\theta)$ $\left|\dfrac{4\cos\theta + \sin\theta - 1}{\sqrt{\cos^2\theta + \sin^2\theta}}\right| = r$ $\Rightarrow\ 4\cos\theta + \sin\theta = 1 - r$ $Q = A + r(\cos\theta,\sin\theta) = (4 + r\cos\theta,\ 1 + r\sin\theta)$ $M = \left(\dfrac{4 + (1+r)\cos\theta}{2},\ \dfrac{1 + (1+r)\sin\theta}{2}\right)$ $4M_x + M_y = \dfrac{17 + (1+r)(4\cos\theta + \sin\theta)}{2}$ $= \dfrac{17 + (1+r)(1-r)}{2} = \dfrac{18 - r^2}{2}$ $\text{line of midpoints}:\ 4x + y = \dfrac{18 - r^2}{2}$ $y = 0:\ 4x = \dfrac{18 - r^2}{2} \Rightarrow x = \dfrac{18 - r^2}{8}$ $B = \left(\dfrac{18 - r^2}{8},\ 0\right)$ $AB^2 = \left(4 - \dfrac{18 - r^2}{8}\right)^2 + 1 = 5$ $\Rightarrow\ \left(4 - \dfrac{18 - r^2}{8}\right)^2 = 4$ $4 - \dfrac{18 - r^2}{8} = \pm 2$ $+2:\ 18 - r^2 = 16 \ \Rightarrow \ r^2 = 2$ $-2:\ 18 - r^2 = 48 \ \Rightarrow \ r^2 = -30 \ \text{(rejected)}$ $\boxed{r^2 = 2}$