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.

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