If (x − a)2 + (y − b)2 = c2, for some c > 0 and y > b, then the expression \(\displaystyle\frac{\left[1+\left(\frac{dy}{dx}\right)^2\right]^{\frac{3}{2}}}{\frac{d^2y}{dx^2}}\) is equal to:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → −c
Differentiating the equation of the circle with respect to x (implicit differentiation):
\( 2(x-a) + 2(y-b)\,y' = 0 \)
\( \displaystyle y' = -\frac{x-a}{y-b} \)
Then
\( \displaystyle 1 + (y')^2 = \frac{(y-b)^2 + (x-a)^2}{(y-b)^2} = \frac{c^2}{(y-b)^2} \)
Differentiating \((x-a) + (y-b)\,y' = 0\) once more:
\( 1 + (y')^2 + (y-b)\,y'' = 0 \)
\( \displaystyle y'' = -\frac{1+(y')^2}{y-b} = -\frac{c^2}{(y-b)^3} \)
Since y > b, the quantity y − b is positive, so
\( \displaystyle \left[1+(y')^2\right]^{3/2} = \left[\frac{c^2}{(y-b)^2}\right]^{3/2} = \frac{c^3}{(y-b)^3} \)
Dividing:
\( \displaystyle \frac{\left[1+(y')^2\right]^{3/2}}{y''} = \frac{c^3}{(y-b)^3}\times\frac{(y-b)^3}{-c^2} \)
\( = -c \)
The value is a constant, independent of a and b.