Question

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 N 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 N is also a common tangent to $C_2$ and C, then the radius of the circle C is _____.

Answer 1

8

Answer 2

Let $O_1$ and $O_2$ be the centers of circles $C_1$ and $C_2$, respectively. Given $r_1 = r_2 = 1$ and the distance between centers $O_1O_2 = 6$. Let N be the midpoint of $O_1O_2$. Thus, $O_1N = NO_2 = 3$.

Let O be the center of circle C and $r$ be its radius. Since C touches $C_1$ and $C_2$ externally, $OO_1 = r+1$ and $OO_2 = r+1$. This implies O lies on the perpendicular bisector of $O_1O_2$, which passes through N.

Set up a coordinate system with N at the origin (0,0), $O_1 = (-3, 0)$ and $O_2 = (3, 0)$. The center O of circle C is on the y-axis, so $O = (0, y_0)$.

The distance $OO_1^2 = (0 - (-3))^2 + (y_0 - 0)^2 = 9 + y_0^2$. Since $OO_1 = r+1$, we have $(r+1)^2 = 9 + y_0^2$.

The common tangent to $C_1$ and $C_2$ passing through N is a transverse common tangent. Let its equation be $y = mx$. The distance from $O_1(-3,0)$ to $mx - y = 0$ is $r_1=1$: $\frac{|m(-3) - 0|}{\sqrt{m^2 + (-1)^2}} = 1 \implies \frac{3|m|}{\sqrt{m^2+1}} = 1 \implies 9m^2 = m^2+1 \implies 8m^2 = 1 \implies m^2 = \frac{1}{8}$.

This tangent is also common to $C_2$ and C. The distance from $O(0, y_0)$ to $mx - y = 0$ is $r$: $\frac{|m(0) - y_0|}{\sqrt{m^2 + (-1)^2}} = r \implies \frac{|y_0|}{\sqrt{m^2+1}} = r$. Substituting $m^2 = 1/8$, we get $\sqrt{m^2+1} = \sqrt{1/8 + 1} = \sqrt{9/8} = 3/\sqrt{8}$. So, $r = \frac{|y_0|}{3/\sqrt{8}} \implies |y_0| = \frac{3r}{\sqrt{8}}$. Squaring both sides: $y_0^2 = \frac{9r^2}{8}$.

Substitute $y_0^2$ into the equation $(r+1)^2 = 9 + y_0^2$: $(r+1)^2 = 9 + \frac{9r^2}{8}$ $r^2 + 2r + 1 = 9 + \frac{9r^2}{8}$ Multiply by 8: $8r^2 + 16r + 8 = 72 + 9r^2$ Rearrange: $r^2 - 16r + 64 = 0$ $(r-8)^2 = 0 \implies r = 8$.