Let the centres of the unit circles $C_1$ and $C_2$ be $A$ and $B$ respectively, with $AB=6.$ Since $P$ is the midpoint of $AB$, we have $AP=PB=3.$

We are told that there is a line passing through $P$ which is:

1. a common tangent to $C_1$ and $C_2$, and
2. also a common tangent to $C_2$ and the circle $C$.

We need the radius of circle $C$, which touches both $C_1$ and $C_2$ externally.

1. Set up coordinates

Take $A(-3,0),\quad B(3,0),\quad P(0,0).$ The circles $C_1$ and $C_2$ have radius $1$.

Let the common tangent through $P$ be $y=mx.$ In general form, $mx-y=0.$

For this line to be tangent to $C_1$ (center $A(-3,0)$, radius $1$), the perpendicular distance from $A$ to the line must be $1$: $\frac{|m(-3)-0|}{\sqrt{m^2+1}}=1.$ So, $\frac{3|m|}{\sqrt{m^2+1}}=1.$ Squaring, $\frac{9m^2}{m^2+1}=1$ $9m^2=m^2+1$ $8m^2=1$ $m^2=\frac18.$ Thus the tangent through $P$ is one of the two lines $y=\pm \frac{x}{2\sqrt2}.$

2. Use the fact that the same line is tangent to $C_2$ and $C$

Since the line is also tangent to $C_2$ and $C$, and $C_2$ has center $B(3,0)$ and radius $1$, the center of circle $C$ must lie on the angle bisector of the tangent configuration on the same side, and because $C$ touches $C_2$ externally, the distance between their centers is $1+r,$ where $r$ is the radius of $C$.

Also, if a line is tangent to a circle, the perpendicular distance from the center to the tangent equals the radius. So if the center of $C$ is $O$, then its distance from the line $y=mx$ equals $r$.

Now circle $C$ touches both $C_1$ and $C_2$ externally. Since $C_1$ and $C_2$ are symmetric about the $y$-axis, the center $O$ of circle $C$ must lie on the perpendicular bisector of $AB$, i.e. on the $y$-axis. So let $O=(0,k).$

Because $OA=OB$ and external tangency gives $OA=OB=r+1.$ Now $OA=\sqrt{(0+3)^2+k^2}=\sqrt{9+k^2}.$ Hence $\sqrt{9+k^2}=r+1. \qquad (1)$

Also, distance of $O=(0,k)$ from the tangent line $mx-y=0$ is $\frac{|m\cdot 0-k|}{\sqrt{m^2+1}}=\frac{|k|}{\sqrt{m^2+1}}=r. \qquad (2)$

Using $m^2=\frac18$, $m^2+1=\frac18+1=\frac98,$ so $\sqrt{m^2+1}=\frac{3}{2\sqrt2}.$ Thus from (2), $r=|k|\cdot \frac{2\sqrt2}{3}. \qquad (3)$

Take $k>0$ (the other tangent gives symmetric situation), so $r=\frac{2\sqrt2}{3}k$ $k=\frac{3r}{2\sqrt2}. \qquad (4)$

Substitute (4) into (1): $\sqrt{9+\left(\frac{3r}{2\sqrt2}\right)^2}=r+1.$ Square both sides: $9+\frac{9r^2}{8}=(r+1)^2=r^2+2r+1.$ Multiply by $8$: $72+9r^2=8r^2+16r+8$ $r^2-16r+64=0$ $(r-8)^2=0.$ Therefore, $r=8.$

3. Final answer

The radius of circle $C$ is $\boxed{8}.$

This matches the stored correct answer.