Circle cuts Circle - (5)

Let the unit circles be represented by:

$C1: x^2+y^2=1$ and $C2: (x-k)^2+y^2=1$

They intersect according to:

$x^2+y^2 =1 = x^2 - 2kx + k^2 + y^2 \Rightarrow 0 = -2kx + k^2 \Rightarrow x = \frac{k}{2}$ ;

and if their common chord is equal to half of their radius, then the two endpoints points include $(x,y) = (k/2, \pm 1/4).$ Solving for $k$ on either circle gives: $(k/2)^2 + (\pm 1/4)^2 = 1 \Rightarrow k = \frac{\sqrt{15}}{2}.$ If we now compute the tangent line slopes of $C1$ and $C2$ at one of these endpoints (WLOG, let us use $(\sqrt{15}/4, 1/4)$ ):

$C1: \frac{dy}{dx} = -\frac{x}{\sqrt{1-x^2}} \Rightarrow \frac{dy}{dx}|_{x=\sqrt{15}/4} = -\frac{\sqrt{15}/4}{\sqrt{1 - (\sqrt{15}/4)^2}} = -\sqrt{15} = m_{1};$

$C2: \frac{dy}{dx} = -\frac{x - \sqrt{15}/2}{\sqrt{1-(x - \sqrt{15}/2)^2}} \Rightarrow \frac{dy}{dx}|_{x=\sqrt{15}/4} = -\frac{\sqrt{15}/4 - \sqrt{15}/2}{\sqrt{1 - (\sqrt{15}/4 - \sqrt{15}/2)^2}} = \sqrt{15} = m_{2}$

and if the angle between these tangent lines is $\theta$ , then it can be computed via:

$\tan \theta = |\frac{m_{2}-m_{1}}{1 + m_{1} m_{2}}| = |\frac{2\sqrt{15}}{1 - 15}| = |-\frac{\sqrt{15}}{7}| = \frac{\sqrt{15}}{7}$ ,

or $\cos \theta = \frac{7}{\sqrt{7^2 + (\sqrt{15})^2}} = \frac{7}{8}.$

Hence, $a = 7, b = 8 \Rightarrow a+b = \boxed{15}.$