# Family of circles

Geometry Level 4

The circle $$x^{ 2 }+{ y }^{ 2 }=8$$ is cut by a series of circles all of which pass through the points $$(4,0)$$ and $$(0,6)$$. The radius of the member of the family whose common chord with the circle $$x^{ 2 }+{ y }^{ 2 }=8$$ passes through $$(12,4)$$ is $$R$$. If $$R$$ is of the form $$\sqrt { a }$$, then find $$a$$.

$$a$$ is a square free positive integer.

×