Question 305478
Make the origin the center of your first circle (left hand side).
The center of the middle circle is then ({{{2R}}},{{{0}}}) and the equation of the middle circle is then,
{{{(x-2R)^2+y^2=R^2}}}
The tangent line also starts at ({{{0}}},{{{0}}}) and has has the formula,
{{{y=mx}}}
The tangent line is the base of a right triangle that has a hypotenuse equal to {{{4R}}} and the other side {{{R}}}. The third side is then, by the Pythagorean theorem,
{{{(4R)^2-R^2=S^2}}}
{{{S^2=16R^2-R^2=15R^2}}}
{{{S=sqrt(15)R}}}
The tangent of the angle that the tangent line makes with the x-axis is then equal to opposite side, {{{R}}}, over the adjacent side, {{{sqrt(15)R}}}. This is also the slope you need "m".
{{{y=(R/(sqrt(15)R))*x}}}
{{{y=x/sqrt(15)}}}
{{{y^2=x^2/15}}}
You can then plug that into your circle equation to find the two intersection points. 
{{{(x-2R)^2+y^2=R^2}}}
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Once you have the points, you can use the distance formula to calculate the distance between the points.
{{{D^2=(x1-x2)^2+(y1-y2)^2}}}