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

If (x-a)² + (y-b)² = c², for some c > 0, prove that fra — CBSE Class 12 Mathematics 2019 PYQ

Question

(x-a)^2 + (y-b)^2 = c^2

CBSECBSE Class XII Board 2019Subjective· 4mImportance★★★★★

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For a circle, the radius of curvature is constant and equals the circle’s radius $c$ . The given expression is exactly the radius of curvature, so it simplifies to $c$ , 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

\rho = \frac{\left[1 + \left(\frac{dy}{dx}\right)^2\right]^{3/2}}{\left|\frac{d^2y}{dx^2}\right|}.

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.

Start with the circle equation

(x-a)^2 + (y-b)^2 = c^2.

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

Differentiate implicitly with respect to $x$

2(x-a) + 2(y-b)\,\frac{dy}{dx} = 0.

Divide through by 2:

(x-a) + (y-b)\,\frac{dy}{dx} = 0.

\frac{dy}{dx} = -\frac{x-a}{y-b}.

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

Differentiate again to get the second derivative

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

\frac{d^2y}{dx^2} = -\frac{(1)(y-b) - (x-a)\frac{dy}{dx}}{(y-b)^2}.

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

\frac{d^2y}{dx^2} = -\frac{(y-b) - (x-a)\left(-\frac{x-a}{y-b}\right)}{(y-b)^2} = -\frac{(y-b) + \frac{(x-a)^2}{y-b}}{(y-b)^2}.

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

\frac{d^2y}{dx^2} = -\frac{\frac{(y-b)^2 + (x-a)^2}{y-b}}{(y-b)^2} = -\frac{(x-a)^2 + (y-b)^2}{(y-b)^3}.

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

\frac{d^2y}{dx^2} = -\frac{c^2}{(y-b)^3}.

Note

The second derivative is negative when $y > b$ (top half of circle) and positive when $y < b$ (bottom half). But its absolute value is what matters for the radius of curvature formula.

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

From step 2:

1 + \left(\frac{dy}{dx}\right)^2 = 1 + \frac{(x-a)^2}{(y-b)^2} = \frac{(y-b)^2 + (x-a)^2}{(y-b)^2} = \frac{c^2}{(y-b)^2}.

Assemble the radius of curvature expression

The numerator of the given expression is

\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}.

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 …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2019Set 65/2/14 marks
$(x-a)^2 + (y-b)^2 = c^2$
›Reveal solutionSolution
For a circle, the radius of curvature is constant and equals the circle’s radius $c$ . The given expression is exactly the radius of curvature, so it simplifies to $c$ , 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
$\rho = \frac{\left[1 + \left(\frac{dy}{dx}\right)^2\right]^{3/2}}{\left|\frac{d^2y}{dx^2}\right|}.$
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.

Start with the circle equation

$(x-a)^2 + (y-b)^2 = c^2.$
Here $c$ is the radius, $a$ and $b$ are the centre coordinates. We treat $y$ as a function of $x$ .

Differentiate implicitly with respect to $x$

$2(x-a) + 2(y-b)\,\frac{dy}{dx} = 0.$
Divide through by 2:
$(x-a) + (y-b)\,\frac{dy}{dx} = 0.$
So
$\frac{dy}{dx} = -\frac{x-a}{y-b}.$
This is the slope of the tangent at any point on the circle.

Differentiate again to get the second derivative

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

$\frac{d^2y}{dx^2} = -\frac{(1)(y-b) - (x-a)\frac{dy}{dx}}{(y-b)^2}.$
Substitute $\frac{dy}{dx} = -\frac{x-a}{y-b}$ into the numerator:
$\frac{d^2y}{dx^2} = -\frac{(y-b) - (x-a)\left(-\frac{x-a}{y-b}\right)}{(y-b)^2} = -\frac{(y-b) + \frac{(x-a)^2}{y-b}}{(y-b)^2}.$
Combine the terms in the numerator over a common denominator $y-b$ :
$\frac{d^2y}{dx^2} = -\frac{\frac{(y-b)^2 + (x-a)^2}{y-b}}{(y-b)^2} = -\frac{(x-a)^2 + (y-b)^2}{(y-b)^3}.$
But $(x-a)^2 + (y-b)^2 = c^2$ , so
$\frac{d^2y}{dx^2} = -\frac{c^2}{(y-b)^3}.$
Note

The second derivative is negative when $y > b$ (top half of circle) and positive when $y < b$ (bottom half). But its absolute value is what matters for the radius of curvature formula.

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

From step 2:

$1 + \left(\frac{dy}{dx}\right)^2 = 1 + \frac{(x-a)^2}{(y-b)^2} = \frac{(y-b)^2 + (x-a)^2}{(y-b)^2} = \frac{c^2}{(y-b)^2}.$

Assemble the radius of curvature expression

The numerator of the given expression is

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

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]3/2d2ydx2\frac{\left[1 + \left(\frac{dy}{dx}\right)^2\right]^{3/2}}{\frac{d^2y}{dx^2}}dx2d2y​[1+(dxdy​)2]3/2​ is a constant independent of a and b.

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

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