# The Osculating Circle at a Point on a Curve

Definition: Let $\vec{r}(t) = (x(t), y(t), z(t))$ be a vector-valued function that traces out the smooth curve $C$, and let $P$ be the point on $C$ at $t$. The Osculating Circle at $P$ is the circle that best approximates $C$ at $P$. The osculating circle at $P$:1) Contains the point $P$.2) Has radius $\rho (t) = \frac{1}{\kappa (t)}$.3) Has curvature $\kappa (t)$.4) Shares the same tangent at $P$. |

*If $y = f(x)$ generates a curve $C$, then the osculating circle at $P$ is also defined to be the circle that best approximates $C$ at $P$ by having radius $\rho (x) = \frac{1}{\kappa (x)}$, curvature $\kappa (x)$, and shares the same tangent at $P$.*

Definition: Let $\vec{r}(t) = (x(t), y(t), z(t))$ be a vector-valued function that traces out the smooth curve $C$, and let $P$ be the point on $C$ at $t$. Then Center of Curvature at $P$ is the center of the osculating circle at $P$. |

From the definition of an osculating circle, we can calculate the center of curvature which we will denote by $\vec{r_c}(t)$, by the following formula:

(1)This can easily be seen as $\vec{r}(t)$ takes us to the point $P$, and then we travel $\rho (t)$ (the radius of the osculating circle at $P$) in the direction of the unit normal vector $\hat{N}(t)$.

## Example 1

**Let $\vec{r}(t) = \left ( t, \frac{1}{2}t^2 + 1, 0 \right )$. Find the equation of the osculating circle at $t = 1$.**

We will first simplify this problem down. We should first notice that the graph of $\vec{r}(t)$ represents a plane curve - namely one that lies on the $xy$-plane. Since $x = t$, then $y = \frac{1}{2}t^2 + 1$ implies that $y = \frac{1}{2}x^2 + 1$. So $\vec{r}(t)$ represents the graph of the single variable function $f(x) = \frac{1}{2}x^2 + 1$.

We will first find the curvature of this function by using the formula $\kappa (x) = \frac{ \mid f''(x) \mid}{(1 + (f'(x))^2)^{3/2}}$ and then take the reciprocal to get the radius of curvature, $\rho (x) = \frac{1}{\kappa (x)}$. To use this formula, we must first calculate $f'(x)$ and $f''(x)$. We have that $f'(x) = x$, and $f''(x) = 1$, and so:

(2)Since $x = t$, the given value $t = 1$ corresponds to $x = 1$. Plugging this into the formula above and we have that $\rho (1) = 2^{3/2}$.

We now need to find the center of curvature, i.e, the center of the osculating circle at $t = 1$. To do so, we need to find the unit normal vector $\hat{N}(1)$. To calculate the unit normal vector, we must first find the unit tangent vector $\hat{T}(1)$ and the unit binormal vector $\hat{B}(1)$ and then take the cross product $\hat{N}(1) = \hat{B}(1) \times \hat{T}(1)$.

First let's calculate the unit tangent vector with the formula $\hat{T}(t) = \frac{\vec{r'}(t)}{\| \vec{r'}(t) \|}$. We note that $\vec{r'}(t) = (1, t, 0)$, and $\| \vec{r'}(t) \| = \sqrt{1 + t^2}$. Therefore $\hat{T}(t) = \left ( \frac{1}{\sqrt{1 + t^2}}, \frac{t}{\sqrt{1+t^2}}, 0 \right )$. Plugging in $t = 1$ and we have that $\hat{T}(1) = \left ( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0 \right )$.

Now let's calculate the unit binormal vector with the formula $\hat{B}(t) = \frac{\vec{r'}(t) \times \vec{r''}(t)}{\| \vec{r'}(t) \times \vec{r''}(t) \|}$. We note that $\vec{r'}(t) = (1, t, 0)$ and $\vec{r''}(t) = (0, 1, 0)$. Therefore $\vec{r'}(t) \times \vec{r''}(t) = (0, 0, 1)$. Clearly $\| \vec{r'}(t) \times \vec{r''}(t) \| = 1$, so $\hat{B}(t) = (0, 0, 1)$. Plugging in $t = 1$ and we get that $\hat{B}(1) = (0, 0, 1)$.

Therefore the unit normal vector can be obtained as $\hat{N}(1) = \hat{B}(1) \times \hat{T}(1)$:

(3)Therefore the center of curvature (the center of the circle) at $t = 1$ is:

(4)In $\mathbb{R}^2$, the center of the osculating circle at $t = 1$ is thus $\left ( -1, \frac{7}{2} \right )$. We calculated the radius of the circle above ($\rho (1) = 2^{3/2}$) and so the equation of the osculating circle is:

(5)In $\mathbb{R}^3$, we can represent this circle with the set of parametric equations $\left\{\begin{matrix}x = 2^{3/2} \cos t - 1\\ y = 2^{3/2} \sin t + \frac{7}{2} \\ z = 0\\ \end{matrix}\right.$.