Q.If $(x-a)^2 + (y-b)^2 = c^2$ for some $c > 0$, then prove that $\dfrac{\left[1+\left(\dfrac{dy}{dx}\right)^2\right]^{3/2}}{\dfrac{d^2y}{dx^2}}$ is independent of $a$ and $b$.

The expression you’re asked about is the radius of curvature of a plane curve. For any curve $y = f(x)$, the radius of curvature $\rho$ at a point is given by

Your problem gives the absolute value version (without modulus, but since $c>0$, the denominator will be positive for a circle — we’ll see why). So the task is: show that for a circle, this quantity is constant and equals $c$, regardless of where the centre $(a,b)$ is.

The intuition: a circle has the same curvature at every point — it bends uniformly. So its radius of curvature is just its radius. The centre coordinates $(a,b)$ only shift the circle around; they don’t change its shape or size. So the result should indeed be independent of $a$ and $b$.

Let’s verify this by direct differentiation.

1. Start with the circle equation

Here $c$ is the radius, $a$ and $b$ are the centre coordinates. We treat $y$ as a function of $x$.

2. Differentiate implicitly with respect to $x$

Divide through by 2:

So

This is the slope of the tangent at any point on the circle.

3. Differentiate again to get the second derivative

   Differentiate $\frac{dy}{dx} = -\frac{x-a}{y-b}$ with respect to $x$. Use the quotient rule:

Substitute $\frac{dy}{dx} = -\frac{x-a}{y-b}$ into the numerator:

Combine the terms in the numerator over a common denominator $y-b$:

But $(x-a)^2 + (y-b)^2 = c^2$, so

4. Now compute $1 + \left(\frac{dy}{dx}\right)^2$

   From step 2:

5. Assemble the radius of curvature expression

   The numerator of the given expression is

The denominator is $\frac{d^2y}{dx^2}$. But careful: the formula uses the absolute value of the second derivative. Since $c>0$, we have …