Radius of Curvature — the Circle that Best Fits a Curve
At any point on a smooth curve, zoom in close enough and the curve looks like a small arc of some circle. That circle — the one that hugs the curve most snugly at the point — is called the osculating circle, and its radius is the radius of curvatureρ at that point. A tightly bending curve is matched by a small circle (small ρ); a nearly straight curve is matched by a huge circle (large ρ).
Curvature and Its Radius
Curvatureκ measures how sharply a curve bends. The radius of curvature is simply its reciprocal:
ρ=κ1.
So more bending ⇒ bigger κ⇒ smaller ρ.
Important
A circle of radius r bends by the same amount at every point, so its curvature is the constant κ=r1 and its radius of curvature is ρ=r everywhere. The circle is the one curve whose curvature radius equals its own radius — which is exactly why it is the yardstick for measuring the bending of every other curve.
The Working Formula
For a curve y=f(x), the radius of curvature at a point is
ρ=dx2d2y[1+(dxdy)2]3/2.
The first derivative fixes the direction (the tangent), and the second derivative supplies the bending; combining them gives the size of the best-fitting circle.
Two Anchoring Cases
A straight line:dx2d2y=0, so ρ→∞ — no bending, an "infinitely large" circle. Sensible. …
For the circle (x−a)2+(y−b)2=c2 the expression is the radius-of-curvature formula and evaluates to ±c (magnitude c), a constant with no a or b in it.
The quantity y′′[1+(y′)2]3/2 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.
Method: Radius of Curvature — Show an Expression Is Independent of the Circle's Centre
The expression y′′[1+(y′)2]3/2 is the standard radius of curvature formula for a plane curve. For a circle, every point has the same curvature, so proving this expression is constant (and, further, that it contains no a or b) is really proving the formula correctly reduces to the circle's own radius c.
Steps
Step 1: Differentiate the circle's equation implicitly to find y1=dxdy in terms of x−a and y−b
Step 2: Use the circle's own equation to simplify 1+(y1)2 into a clean expression involving only c and (y−b)
Step 3: Differentiate y1 again (quotient rule), substituting the circle's equation once more to simplify y2
Step 4: Assemble the full ratio and confirm that a and b cancel out entirely, leaving only c …
Mistake 1: Not substituting the circle's own equation back in when simplifying 1+(y1)2 and y2.
Why it's wrong: without recognising that (x−a)2+(y−b)2=c2 can replace a messier expression at two separate points in the derivation, the algebra never collapses to something free of a and b — the whole point of the proof is lost. Correct approach: actively look for opportunities to re-use the given circle equation while simplifying, not just at the very start.
Mistake 2: Forgetting to track the sign/absolute-value subtlety when taking the 3/2 power of (y−b)2c2. …
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
Q.If (x−a)2+(y−b)2=c2, for some c>0, prove that dx2d2y[1+(dxdy)2]3/2 is a constant independent of a and b.
›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 ρ at a point is given by
ρ=dx2d2y[1+(dxdy)2]3/2.
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=c2.
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)dxdy=0.
Divide through by 2:
(x−a)+(y−b)dxdy=0.
So
dxdy=−y−bx−a.
This is the slope of the tangent at any point on the circle.
Differentiate again to get the second derivative
Differentiate dxdy=−y−bx−a with respect to x. Use the quotient rule:
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.