Family of circles

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.