Question

If \((x-c)^{2}+(y-a)^{2}=b^{2}\), then $\frac{\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{3 / 2}}{\frac{d^{2} y}{d x^{2}}}$ is independent of (A) \(\mathrm{b}\) (B) a (C) \(\mathrm{c}\) at \(\mathrm{x}=0\) (D) None of these

Short answer

Answer: (C) \(c\) at \(x=0\).

Step-by-step solution

Differentiate the circle equation with respect to x

Differentiate the equation of the circle with respect to x: \((x-c)^{2}+(y-a)^{2}=b^{2}\) We'll use the chain rule for this task: \(2(x-c) + 2(y-a)\frac{dy}{dx} = 0\) Now, let's solve the equation for \(\frac{dy}{dx}\): \(\frac{dy}{dx} = -\frac{x-c}{y-a}\)

Differentiate the expression for dy/dx with respect to x again

Now we need to find the second derivative, \(\frac{d^2y}{dx^2}\): \(\frac{d^2y}{dx^2} = \frac{d}{dx} \left(-\frac{x-c}{y-a}\right)\) Using the quotient rule of differentiation: \(\frac{d^2y}{dx^2} = \frac{-(y-a) - (x-c)\frac{dy}{dx}}{(y-a)^2}\)

Compute the given expression and analyze its dependence on variables

Let's compute the expression \(\frac{\left[1+\left(\frac{dy}{dx}\right)^{2}\right]^{3/2}}{\frac{d^2y}{dx^2}}\) Using the expressions for \(\frac{dy}{dx}\) and \(\frac{d^2y}{dx^2}\) we found earlier: \(\frac{\left[1+\left(-\frac{x-c}{y-a}\right)^{2}\right]^{3/2}}{\frac{-(y-a)-(x-c)\frac{dy}{dx}}{(y-a)^2}}\) Now, observe the expression and the variables within it: - It clearly depends on \(a\) and \(c\), since we have both terms in the numerator and the denominator of the expression. - We can see that it doesn't depend on \(x\) when \(x=0\) because all terms containing \(x\) will equate to zero in that case. - The expression also depends on \(b\), as it is present in the equation for the circle, which we used to obtain the derivatives. Based on the observation above, we can conclude that the expression is independent of: (C) \(c\) at \(x=0\) So, the correct answer is (C).