The centres of two circles $C_1$ and $C_2$ each of unit radius are at a distance of $6$ units from each other. Let $P$ be the mid point of the line segment joining the centres of $C_1$ and $C_2$ and $C$ be a circle touching circles $C_1$ and $C_2$ externally. If a common tangent to $C_1$ and $C$ passing through $P$ is also a common tangent to $C_2$ and $C$, then the radius of the circle $C$ is

The centres of two circles C₁ and C₂ each of unit radius are at a distance of 6 units from each other. Let P be the mid point of the line segment joining the centres of C₁ and C₂ and C be a circle touching circles C₁ and C₂ externally. If a common tangent to C₁ and C passing through P is also a common tangent to C₂ and C, then the radius of the circle C is $\text{Given: } r_1 = r_2 = 1,\qquad O_1O_2 = 6$ $P = \text{midpoint of } O_1O_2 \Rightarrow O_1P = PO_2 = 3$ $C \text{ touches } C_1 \text{ and } C_2 \text{ externally}$ $\text{To find: radius } r \text{ of circle } C$ $\text{A common tangent to } C_1, C \text{ through } P$ $\text{is also a common tangent to } C_2, C$ $\text{Plan: } 1.\ \text{coordinates} \quad 2.\ O(0,k),\ r$ $3.\ \text{tangency equations} \quad 4.\ \text{solve for } r$ $O_1(-3,0),\quad O_2(3,0),\quad P(0,0)$ $r_1 = r_2 = 1$ $O(0,k)\ \text{on the perpendicular bisector of } O_1O_2$ $\text{radius of } C = r$ $OO_1 = r + r_1 = r+1$ $\sqrt{9+k^2} = r+1 \quad (1)$ $\text{tangent through } P:\quad y = mx$ $\text{dist}(O_1,\ y=mx) = \dfrac{|-3m|}{\sqrt{1+m^2}} = 1$ $9m^2 = 1+m^2 \ \Rightarrow \ m^2 = \dfrac{1}{8}$ $1+m^2 = \dfrac{9}{8},\qquad \sqrt{1+m^2} = \dfrac{3}{2\sqrt{2}} \quad (2)$ $\dfrac{k}{\sqrt{1+m^2}} = r \ \Rightarrow \ r = \dfrac{2\sqrt{2}}{3}\,k \quad (3)$ $\sqrt{9+k^2} = 1 + \dfrac{2\sqrt{2}}{3}\,k$ $9+k^2 = 1 + \dfrac{4\sqrt{2}}{3}k + \dfrac{8}{9}k^2$ $k^2 - 12\sqrt{2}\,k + 72 = 0$ $\Delta = (12\sqrt{2})^2 - 4\cdot 1\cdot 72 = 288-288 = 0$ $(k-6\sqrt{2})^2 = 0 \ \Rightarrow \ k = 6\sqrt{2}$ $r = \dfrac{2\sqrt{2}}{3}\cdot 6\sqrt{2} = 8$ $\sqrt{9+72} = 9 = r+1$ $r = 8$