Radius of Externally Touching Circle with Common Tangent

To find the radius of the circle $C$, let us set up a Cartesian coordinate system:

1. Coordinate System and Centres:
   Let the midpoint $P$ of the line segment joining the centres of circles $C_1$ and $C_2$ be the origin $(0, 0)$.
   Since the distance between the centres of $C_1$ and $C_2$ is $6$ units, the line joining their centres can be taken as the $x$-axis.

   • Centre of $C_1$: $O_1 = (-3, 0)$
   • Centre of $C_2$: $O_2 = (3, 0)$
   • The radius of each circle is $r_1 = r_2 = 1$.

2. Properties of Circle $C$:
   Let the radius of the circle $C$ be $R$, and let its centre be $O$.
   Since $C$ touches both $C_1$ and $C_2$ externally, the distances from $O$ to $O_1$ and $O_2$ are:
   $O O_1 = O O_2 = R + 1$
   Since $O$ is equidistant from $O_1$ and $O_2$, it lies on the perpendicular bisector of $O_1 O_2$, which is the $y$-axis.
   Thus, the centre of $C$ is $O = (0, y_0)$. Without loss of generality, assume $y_0 > 0$.
   Using the distance formula from $O(0, y_0)$ to $O_1(-3, 0)$:
   $O O_1^2 = 3^2 + y_0^2 = (R + 1)^2$
   $y_0^2 = (R + 1)^2 - 9 \quad \text{--- (1)}$

3. Condition for the Common Tangent:
   A common tangent passing through $P(0, 0)$ is tangent to $C_1$, $C_2$, and $C$.
   Let the equation of this tangent line $L$ passing through $(0, 0)$ be:
   $mx - y = 0$
   Since $L$ is tangent to $C_2$ (which has centre $(3, 0)$ and radius $1$):
   $\frac{|3m - 0|}{\sqrt{m^2 + 1}} = 1 \implies 9m^2 = m^2 + 1 \implies 8m^2 = 1 \implies m^2 = \frac{1}{8}$
   From this, we have:
   $\sqrt{m^2 + 1} = \sqrt{\frac{1}{8} + 1} = \sqrt{\frac{9}{8}} = \frac{3}{2\sqrt{2}}$

   Since $L$ is also tangent to circle $C$ (which has centre $(0, y_0)$ and radius $R$):
   $\frac{|m(0) - y_0|}{\sqrt{m^2 + 1}} = R \implies \frac{y_0}{\sqrt{m^2 + 1}} = R$
   $y_0 = R \sqrt{m^2 + 1} = R \sqrt{\frac{9}{8}}$
   Squaring both sides:
   $y_0^2 = \frac{9}{8} R^2 \quad \text{--- (2)}$

4. Solving for $R$:
   Equating equations (1) and (2):
   $(R + 1)^2 - 9 = \frac{9}{8} R^2$
   $R^2 + 2R + 1 - 9 = \frac{9}{8} R^2$
   $R^2 + 2R - 8 = \frac{9}{8} R^2$
   Multiplying through by $8$:
   $8R^2 + 16R - 64 = 9R^2$
   $R^2 - 16R + 64 = 0$
   $(R - 8)^2 = 0 \implies R = 8$

Thus, the radius of the circle $C$ is $8$.