If the tangent at $$\left( {1,\,7} \right)$$  to the curve $${x^2} = y - 6$$   touches the circle $${x^2} + {y^2} + 16x + 12y + c = 0$$       then the value of $$c$$ is :

Solution :
Equation of tangent at $$\left( {1,\,7} \right)$$  to $${x^2} = y - 6$$   is $$2x - y + 5 = 0.$$

Now, perpendicular from centre $$O\left( { - 8,\, - 6} \right)$$   to $$2x-y+5=0$$    should be equal to radius of the circle
$$\eqalign{ & \therefore \left| {\frac{{ - 16 + 6 + 5}}{{\sqrt 5 }}} \right| = \sqrt {64 + 36 - C} \cr & \Rightarrow \sqrt 5 = \sqrt {100 - c} \cr & \Rightarrow c = 95 \cr} $$