If the equation of pair of direct common tangent to the circles ${ \left( x+2 \right) }^{ 2 }+{ y }^{ 2 }=1$ and ${ \left( x-4 \right) }^{ 2 }+{ y }^{ 2 }=9$ is given by ${ x }^{ 2 }+2hxy+b{ y }^{ 2 }+2gx+2fy+c=0$. Then ${ b }^{ 2 }-c$ is equal to

The direct common tangents to circles (x+2)^2 + y^2 = 1 and (x-4)^2 + y^2 = 9. The centers are (-2, 0) and (4, 0) with radii 1 and 3. The external center of similitude divides the centers in ratio 1:3 externally: ((1*4 - 3*-2)/(1-3), 0) = (10/-2, 0) = (-5, 0). The tangent lines pass through (-5, 0). Using the condition for tangency, the equation leads to b^2 - c = 39.