Official solution
x2 + y2 + 4x - 6y - 12 = 0
Equation of tangent at (1, - 1)
x - y + 2(x + 1) - 3(y - 1) - 12 = 0
3x - 4y - 7 = 0
\[ \therefore \] Equation of circle is
(x2 + y2 + 4x - 6y - 12) + \[\lambda \] (3x - 4y - 7) = 0
It passes through (4, 0) :
(16 + 16 - 12) + \[\lambda \] (12 - 7) = 0
\[ \Rightarrow \] 20 + \[\lambda (5) = 0
\[ \Rightarrow \] \[\lambda \] = - 4
\[ \therefore \] (x2 + y2 + 4x - 6y - 12) - 4(3x - 4y - 7) = 0
or x2 + y2 - 8x + 10y + 16 = 0
Radius = \[\sqrt {16 + 25 - 16} = 5\]