If $${\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {c^2},$$      for some $$c > 0$$  then $$\frac{{{{\left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right]}^{\frac{3}{2}}}}}{{\frac{{{d^2}y}}{{d{x^2}}}}}$$    is :

Solution :
$$\eqalign{ & {\text{Given relation is }}{\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {c^2},\,c > 0 \cr & {\text{let }}x - a = c\,\cos \,\theta {\text{ and }}y - b = c\,\sin \,\theta \cr & {\text{Therefore,}} \cr & \frac{{dx}}{{d\theta }} = - c\,\sin \,\theta {\text{ and }}\frac{{dy}}{{d\theta }} = c\,\cos \,\theta \cr & \therefore \,\frac{{dy}}{{dx}} = - \cot \,\theta \cr & {\text{Differentiable both sides with respect to}}\,\theta {\text{, we get}} \cr & \frac{d}{{d\theta }}\left( {\frac{{dy}}{{dx}}} \right) = \frac{d}{{d\theta }}\left( { - \cot \,\theta } \right) \cr & {\text{or, }}\frac{d}{{dx}}\left( {\frac{{dy}}{{dx}}} \right)\frac{{dx}}{{d\theta }} = {\text{cose}}{{\text{c}}^2}\theta \cr & {\text{or, }}\frac{{{d^2}y}}{{d{x^2}}}\left( { - c\,\sin \,\theta } \right) = {\text{cose}}{{\text{c}}^2}\theta \cr & {\text{or, }}\frac{{{d^2}y}}{{d{x^2}}} = - \frac{{{\text{cose}}{{\text{c}}^2}\theta }}{c} \cr & \therefore \,\frac{{{{\left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right]}^{\frac{3}{2}}}}}{{\frac{{{d^2}y}}{{d{x^2}}}}} = \frac{{c{{\left[ {1 + {{\cot }^2}\theta } \right]}^{\frac{3}{2}}}}}{{ - {\text{cose}}{{\text{c}}^3}\theta }} = \frac{{c{{\left( {{\text{cose}}{{\text{c}}^2}\theta } \right)}^{\frac{3}{2}}}}}{{ - {\text{cose}}{{\text{c}}^3}\theta }} = - c \cr & {\text{which is constant and is independent of }}a{\text{ and }}b. \cr} $$