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

Question

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

Accepted Answer

> [!TLDR]
> For the circle $(x-a)^2+(y-b)^2=c^2$ the expression is the radius-of-curvature formula and evaluates to $\pm c$ (magnitude $c$), a constant with no $a$ or $b$ in it.

The quantity $\dfrac{\left[1+(y')^2\right]^{3/2}}{y''}$ is exactly the radius of curvature of a plane curve. For a circle the curvature is the same at every point, so this should collapse to the radius $c$. Let us prove it by direct differentiation.

## Step 1 — first derivative
Differentiate $(x-a)^2+(y-b)^2=c^2$ implicitly:
$$2(x-a)+2(y-b)\frac{dy}{dx}=0 \;\Rightarrow\; \frac{dy}{dx} = -\frac{x-a}{y-b}.$$

## Step 2 — the numerator factor
$$1+\left(\frac{dy}{dx}\right)^2 = 1 + \frac{(x-a)^2}{(y-b)^2} = \frac{(x-a)^2+(y-b)^2}{(y-b)^2} = \frac{c^2}{(y-b)^2}.$$

## Step 3 — second derivative
Using the quotient rule on $\frac{dy}{dx}=-\frac{x-a}{y-b}$:
$$\frac{d^2y}{dx^2} = -\frac{(y-b) - (x-a)\frac{dy}{dx}}{(y-b)^2}.$$
Substitute $\frac{dy}{dx}=-\frac{x-a}{y-b}$:
$$\frac{d^2y}{dx^2} = -\frac{(y-b)+\frac{(x-a)^2}{y-b}}{(y-b)^2} = -\frac{(x-a)^2+(y-b)^2}{(y-b)^3} = -\frac{c^2}{(y-b)^3},$$
where the last step uses the circle equation.

## Step 4 — assemble the ratio
$$\left[1+\left(\frac{dy}{dx}\right)^2\right]^{3/2} = \left(\frac{c^2}{(y-b)^2}\right)^{3/2} = \frac{c^3}{|y-b|^3},$$
so …

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

Concept understanding — Circle Curvature Radius

Radius of Curvature — the Circle that Best Fits a Curve

Curvature and Its Radius

The Working Formula

Two Anchoring Cases

Step 1 — first derivative

Step 2 — the numerator factor

Step 3 — second derivative

Step 4 — assemble the ratio

Method: Radius of Curvature — Show an Expression Is Independent of the Circle's Centre

Steps

Common Mistakes

Q.If (x−a)2+(y−b)2=c2(x-a)^2 + (y-b)^2 = c^2(x−a)2+(y−b)2=c2, for some c>0c > 0c>0, prove that [1+(dydx)2]32d2ydx2\frac{\left[1+\left(\frac{dy}{dx}\right)^2\right]^{\frac{3}{2}}}{\frac{d^2y}{dx^2}}dx2d2y​[1+(dxdy​)2]23​​ is a constant independent of aaa and bbb.

Step 1 — first derivative

Step 2 — the numerator factor

Step 3 — second derivative

Step 4 — assemble the ratio

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

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