If (x−1)2+(y−b)2=c2, for some c>0, prove that [1+(dydx)2]32d2ydx2 – InterviewSolution

1.	If (x−1)2+(y−b)2=c2, for some c>0, prove that [1+(dydx)2]32d2ydx2 is a constant independent of a and b.
Answer» $I f (x - 1)^{2} + (y - b)^{2} = c^{2}, f o r s o m e c > 0, p r o v e t h a t \frac{[1 + {(\frac{d y}{d x})}^{2}]^{\frac{3}{2}}}{\frac{d^{2} y}{d x^{2}}} i s a c o n s t a n t i n d e p e n d e n t o f a a n d b .$

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