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Osculating Circle — Definition, Formula & Examples

An osculating circle is the circle that hugs a curve most closely at one point. It touches the curve there, bends the same way, and bends by the same amount.

Let r(t)\mathbf{r}(t) be a smooth curve with curvature κ>0\kappa > 0 at t0t_0. The osculating circle at r(t0)\mathbf{r}(t_0) lies in the osculating plane, which is spanned by the unit tangent T\mathbf{T} and the principal unit normal N\mathbf{N}. Its radius is R=1/κR = 1/\kappa, and its center is c=r(t0)+R N(t0)\mathbf{c} = \mathbf{r}(t_0) + R\,\mathbf{N}(t_0). It agrees with the curve to second order: same point, same tangent line, same curvature.

Key Formula

κ=∥r′(t)×r′′(t)∥∥r′(t)∥3,R=1κ,c=r(t)+R N(t)\kappa = \frac{\lVert \mathbf{r}'(t) \times \mathbf{r}''(t) \rVert}{\lVert \mathbf{r}'(t) \rVert^{3}}, \qquad R = \frac{1}{\kappa}, \qquad \mathbf{c} = \mathbf{r}(t) + R\,\mathbf{N}(t)
Where:
  • r(t)\mathbf{r}(t) = Position vector of the curve
  • κ\kappa = Curvature at the point (must be nonzero)
  • RR = Radius of curvature, the radius of the osculating circle
  • N(t)\mathbf{N}(t) = Principal unit normal, pointing toward the concave side of the curve
  • c\mathbf{c} = Center of curvature, the center of the osculating circle

Worked Example

Problem: Find the osculating circle of the parabola r(t)=⟨t, t2, 0⟩\mathbf{r}(t) = \langle t,\, t^2,\, 0 \rangle at the origin (t=0t = 0).
Step 1: Differentiate twice and evaluate at t=0t = 0.
r′(t)=⟨1,2t,0⟩⇒r′(0)=⟨1,0,0⟩,r′′(0)=⟨0,2,0⟩\mathbf{r}'(t) = \langle 1, 2t, 0 \rangle \Rightarrow \mathbf{r}'(0) = \langle 1, 0, 0 \rangle, \qquad \mathbf{r}''(0) = \langle 0, 2, 0 \rangle
Step 2: Compute the curvature with the cross product. Here ∥r′(0)∥=1\lVert \mathbf{r}'(0) \rVert = 1.
r′(0)×r′′(0)=⟨0,0,2⟩,κ=213=2\mathbf{r}'(0) \times \mathbf{r}''(0) = \langle 0, 0, 2 \rangle, \qquad \kappa = \frac{2}{1^3} = 2
Step 3: The radius is 1/κ1/\kappa. The parabola opens upward, so N(0)=⟨0,1,0⟩\mathbf{N}(0) = \langle 0, 1, 0 \rangle. Move a distance RR from the origin along N\mathbf{N} to find the center.
R=12,c=⟨0,0,0⟩+12⟨0,1,0⟩=⟨0,12,0⟩R = \frac{1}{2}, \qquad \mathbf{c} = \langle 0,0,0 \rangle + \tfrac{1}{2}\langle 0,1,0 \rangle = \left\langle 0, \tfrac{1}{2}, 0 \right\rangle
Answer: The osculating circle has radius 12\tfrac{1}{2} and center (0,12)\left(0, \tfrac{1}{2}\right) in the xyxy-plane. Its equation is x2+(y−12)2=14x^2 + \left(y - \tfrac{1}{2}\right)^2 = \tfrac{1}{4}.

Why It Matters

In multivariable calculus and physics, the osculating circle explains centripetal acceleration. A car on a curved road feels acceleration v2/Rv^2/R, where RR is the local radius of curvature. Highway and railway engineers use the same radius to set safe speed limits and banking angles on curves.

Common Mistakes

Mistake: Placing the center a distance κ\kappa from the point, or moving along the tangent direction.
Correction: The center sits a distance R=1/κR = 1/\kappa from the point along the unit normal N\mathbf{N}, on the concave side. A sharper bend means a larger κ\kappa and a smaller circle.

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