Realising that, #(x-a)^2+(y-b)^2=c^2# represents a Circle with
Centre #(a,b)# and Radius #c#, let us use its
parametric eqns. :
#x=a+c*cos t, y=b+c*sint; t in [0,2pi)" (preferably)"#.
Then, we have, #dx/dt=-c*sint, dy/dt=c*cost#,
so that, #dy/dx=-cott...(1)#.
#:. (d^2y)/dx^2=d/dx{dy/dx}=d/dx{-cott}#,
#=d/dt{-cott}*dt/dx.........[because," the Chain Rule]"#,
#=-(-csc^2t)*{1/(dx/dt)}...[because, dt/dx=1/(dx/dt)]#,
#=csc^2t/(-c*sint)#.
#rArr (d^2y)/dx^2=-1/c*csc^3t......................(2)#.
Hence, by #(1) and (2),#
#{1+(dy/dx)^2}^(3/2)/{(d^2y)/dx^2}#,
#={1+(-cott)^2}^(3/2)/{-1/c*csc^3t}#,
#=-c*(csc^2t)^(3/2)/csc^3t#,
#=-c," a constant, as desired!"#
#"BONUS : "#
For the curve #y=f(x), kappa={(d^2y)/dx^2}/{1+(dy/dx)^2}^(3/2)#,
is called the curvature of the curve, and, #1/kappa#, its radius.
Feel & Spread the Joy of Maths.!