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

Concept: Circle Curvature Radius — the expression given is the radius of curvature of a plane curve, and for a circle it equals the circle’s radius $c$, a constant.

Step 1 — Differentiate the circle equation implicitly:

Step 2 — Differentiate again (using quotient rule):

Substitute $\frac{dy}{dx}$ from Step 1:

Step 3 — Compute $1 + \left(\frac{dy}{dx}\right)^2$:

Step 4 — Form the given expression: …

We are given the equation of a circle: $(x - a)^2 + (y - b)^2 = c^2$, with $c > 0$. The expression

looks intimidating, but it has a beautiful geometric meaning. It is the radius of curvature of the curve at a point — the radius of the circle that best approximates the curve at that point. For a circle itself, the radius of curvature is simply its own radius, everywhere. So the answer should be $c$, a constant independent of $a$ and $b$. Let’s verify this by direct differentiation.

1. Differentiate implicitly the circle equation with respect to $x$:

Divide through by 2:

So

2. Differentiate again to get $\frac{d^2y}{dx^2}$. Use the quotient rule on $\frac{dy}{dx} = -\frac{x - a}{y - b}$:

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

3. Combine the numerator over a common denominator $y - b$:

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

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

5. Plug into the expression: …

Method: Proving a Derivative-Based Expression Is Constant for an Implicitly Defined Family of Curves

Use this method whenever a family of curves is defined implicitly with parameters (like $a, b$ here), and you must show some combination of $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ is a constant that doesn't depend on those parameters.

Steps

Step 1: Differentiate the implicit relation once

Differentiate both sides of the given equation with respect to $x$, treating $y$ as a function of $x$ (chain rule on every $y$-term). Solve the resulting equation for $\dfrac{dy}{dx}$ in terms of $x$, $y$, and the parameters.

Step 2: Differentiate again to get the second derivative

Differentiate the expression for $\dfrac{dy}{dx}$ found in Step 1 with respect to $x$ (typically via the quotient rule), substituting the Step-1 result back in wherever $\dfrac{dy}{dx}$ appears in the new expression. Keep track of signs carefully here — this step is the most error-prone.

Step 3: Substitute into the target expression

Plug both $\dfrac{dy}{dx}$ and $\dfrac{d^2y}{dx^2}$ into the expression the problem asks you to evaluate, and simplify algebraically.

Step 4: Eliminate the parameters using the original implicit relation …

Common Mistakes

Mistake 1: Losing the negative sign in the quotient-rule step

When differentiating $\dfrac{dy}{dx} = -\dfrac{x-a}{y-b}$ a second time, it's easy to drop the leading minus sign or mismanage it partway through the quotient rule. Why it's wrong: the sign of $\dfrac{d^2y}{dx^2}$ directly affects whether the final ratio comes out positive or negative, and a dropped sign silently flips the answer. Correct approach: write the quotient rule out fully with the sign kept as an explicit factor, and re-check it at the end by confirming the sign matches what the geometry predicts (a circle curves the same way at every point).

Mistake 2: Forgetting to substitute the original curve equation back in …