Normal to a Circle- Narlin Beaty

Normal to a Circle in the Plane

Tangent and Normal using Cartesian Coordinates

Disclaimer: Cartesian coordinates in 3d describe surfaces, not curves. A circle is a curve and furthermore, it is always in some plane. In this section we are only going to consider a circle in the $xy$ plane. By using rotation and translation, we could put any circle into the $xy$ plane and then having found the tangent and normal, reverse the operations, applying them to the newly found objects. We aren't going to do that here either.

In Cartesian coordinates our circles is \[ \left(x-h\right)^{2}+\left(y-k\right)^{2}=r^{2} \] The constants, $(h,k)$ are the $x$ and $y$ coordinates of the circle center. The value $r$ is the circle radius. In standard form, the circle equation becomes \[ x^{2}+y^{2}-2hx-2ky=r^{2}-h^{2}-k^{2}. \] For finding the tangent and then the normal (the perpendicular), we usually use derivatives and they might be slightly easier in the standard form.

Tangent using Parametric Coordinates

In parametric coordinates, a circle with radius $R,$ about center point, $C,$ in a plane with basis vectors $\mathbf{u}$ and $\mathbf{v}$ is defined by \begin{equation} \mathbf{X}=C+R\cdot\cos\theta\cdot\mathbf{u}+R\sin\theta\cdot\mathbf{v}.\label{eq:GeneralParametricCircle} \tag{EQ 1} \end{equation} I have used $\mathbf{X}$ to represent either a $2$ or $3$ dimensional vector. $\theta$ is the parameter and spans $2\pi.$ We are considering vector functions of the type $\mathbb{R}\rightarrow\mathbb{R}^{(2\text{ or }3)}$. Our generalized vector function will be named $\mathbf{r(t)}.$ Its first derivative is the tangent. \[ \mathbf{r'}=\left\{ \frac{\partial r_{1}}{\partial t},\frac{\partial r_{2}}{\partial t},\frac{\partial r_{3}}{\partial t}\cdots\right\} \] Although the vector function derivative is general to $n$ dimensions, we are only to consider $\mathbb{R}^{2}$ and $\mathbb{R}^{3}.$ Let $\mathbf{r}$ be the parametric circle of \ref{eq:GeneralParametricCircle}. We need to identify $r_{1}(t),$$r_{2}(t)$ and $r_{3}(t).$ \[ \mathbf{r}=\left(\begin{array}{c} r_{1}\\ r_{2}\\ r_{3} \end{array}\right)=\left(\begin{array}{c} C_{x}\\ C_{y}\\ C_{z} \end{array}\right)+R\cdot\cos\theta\cdot\left(\begin{array}{c} u_{x}\\ u_{y}\\ u_{z} \end{array}\right)+R\cdot\sin\theta\cdot\left(\begin{array}{c} v_{x}\\ v_{y}\\ v_{z} \end{array}\right) \] \[ \mathbf{r^{\prime}=}\left(\begin{array}{c} r_{1}^{\prime}\\ r_{2}^{\prime}\\ r_{3}^{\prime} \end{array}\right)=-R\cdot\sin\theta\cdot\left(\begin{array}{c} u_{x}\\ u_{y}\\ u_{z} \end{array}\right)+R\cdot\cos\theta\cdot\left(\begin{array}{c} v_{x}\\ v_{y}\\ v_{z} \end{array}\right) \] Once $\theta$ is given a value, $\mathbf{r^{\prime}}$ resolves into a single $3$ element vector. That vector is the direction vector of the tangent line. \[ \mathbf{r^{\prime}=\text{Tangent Line Direction Vector}} \] The tangent line itself is the usual \[ \text{Line(x,y,z)}=P(\theta)+t\mathbf{\cdot r^{\prime}}(\theta)\qquad\left(-10\le t\le10\right) \label{EQ2} \tag{EQ 2} \] Here, $P$ is a point on the circle and $\theta$ is the parameter used to get both the point and the tangent vector. In equation \ref{EQ2}, $P(\theta)$ and $\mathbf{r^{\prime}}(\theta)$ are function notation.

Example:Write the circle equation for $Center:C=(2,2),\ radius:R=2,\ \mathbf{u}=(1,0,0),\ \mathbf{v}=(0,1,0).$ Write the equation of its tangent line for a parameter $q:$ $\left\{ -\pi\le q\le\pi\right\} .$
Answer: We just need to copy the equations above and substitute values in place of variables. Notice that these basis vectors are the $xy$ plane. \[ Circle=\left(\begin{array}{c} 2\\ 2\\ 0 \end{array}\right)+2\cos t\cdot\left(\begin{array}{c} 1\\ 0\\ 0 \end{array}\right)+2\sin t\cdot\left(\begin{array}{c} 0\\ 1\\ 0 \end{array}\right)\qquad-\pi\le t\le\pi\Leftarrow\text{Just draws the circle} \] \[ \mathbf{u=}\left(\begin{array}{c} 1\\ 0\\ 0 \end{array}\right)\qquad\mathbf{v}=\left(\begin{array}{c} 0\\ 1\\ 0 \end{array}\right)\qquad-\pi\le q\le\pi\qquad C=\left(\begin{array}{c} 2\\ 2\\ 0 \end{array}\right) \] \[ \mathbf{r^{\prime}}(q)=-2\sin q\cdot\mathbf{u}+\cos q\cdot\mathbf{v}\quad\Leftarrow\text{a vector} \] \[ P(q)=C+2\cos q\cdot\mathbf{u}+2\sin q\cdot\mathbf{v}\Leftarrow\text{a point on the circle} \] \[ Tangent(q)=P(q)+t\cdot\mathbf{r^{\prime}}(q)\qquad-\infty\le t\le\infty\Leftarrow\text{The tangent line} \]

Normal using Parametric Coordinates

Normal in 2d

The normal is just a perpendicular to the tangent line at the tangent point. If we are in $\mathbb{R}^{2},$ we would do it as usual and just get a ``normal'' vector by swapping the $x$ and $y$ vector coordinates and changing the sign of one of them. So \[ \mathbf{r^{\prime}}=\left(\begin{array}{c} r_{x}^{\prime}\\ r_{y}^{\prime} \end{array}\right) \] then \[ \text{normal vector}=\mathbf{n}=\left(\begin{array}{c} r_{y}^{\prime}\\ -r_{x}^{\prime} \end{array}\right) \] Then, using the parameter $q$ from the prior example, the normal line is then \[ Normal(q)=P(q)+t\cdot\mathbf{n}(q)\qquad-\infty\le t\le\infty\Leftarrow\text{The normal line} \]

Let $Center=(2,2)$ and $R=1.$ Show the normal at some point $A.$ Answer: Circle: $(x-2)^{2}+(y-2)^{2}=1$. In standard form, $x^{2}+y^{2}-4x-4y+7=0.$ $\mathbf{r}^{\prime}=(2,2),$ so $\mathbf{n}=(2,-2).$ Let arbitrary point \[ A=\left(\frac{3}{2},\frac{\sqrt{3}+4}{2}\right). \] Translate the normal vector $(2,-2)$ to point $A.$

Normal to Circle in 3d

Special Method for the Circle

The normal vector for a circle is super easy, although not very general. It is the vector from the center point, $C,$ to any point, $P,$ on the circle. \[ \mathbf{\hat{n}}=\frac{P-C}{|P-C|}=\left(\begin{array}{c} n_{x}\\ n_{y}\\ n_{z} \end{array}\right) \] The normal describes a plane which is tangent to the circle at point $P$. \[ tangent\,\,plane:\ n_{x}x+n_{y}y+n_{z}z=n_{x}P_{x}+n_{y}P_{y}+n_{z}P_{z} \] By picking another point $Q,$on the circle we can do the vector cross product \[ circle\ \ normal\ \ vector:\mathbf{c}=(P-C)\times(Q-C) \] to get the plane of the circle. \[ circle\ \ plane:\ \ c_{x}x+c_{y}y+c_{z}z=c_{x}P_{x}+c_{y}P_{y}+c_{z}P_{z} \] The intersection of these two planes is the tangent line. To get the direction vector of that tangent, we just cross the two plane normals. I am arbitrarily naming the tangent vector $\mathbf{r^{\prime}.}$ \[ Tangent\ \ Direction\ \ Vector:\ \mathbf{r^{\prime}=}(\mathbf{c}\times\mathbf{n)} \] \[ Tangent\ \ Line:\ \mathbf{X}=P+t\cdot\mathbf{r^{\prime}} \]

Tangent to the 3d Circle: The circle in 3d uses any 2 orthogonal unit vectors, a radius $R,$ and a Center. \[ \hat{\mathbf{u}}=\frac{(1,1,2)}{|(1,1,2)|}\qquad\hat{\mathbf{v}}=\frac{(1,1,0)}{|(1,1,0)|}\qquad R=2\qquad C=(3,-3,1) \] Circle$=C+R\cos(t)\cdot\hat{\mathbf{u}}+R\sin(t)\cdot\hat{\mathbf{v}}.$ Point $A$ is any point on the circle. The perpendicular to the circle can be named vector $\hat{\mathbf{n}}.$ It is obtained by $\hat{\mathbf{n}}=\frac{(A-C)}{|A-C|}$. It is also the normal to the tangent plane at point $A,$ the yellow plane. The circle lies in the blue plane. Its normal can be found from the cross product, $\hat{\mathbf{w}}=\hat{\mathbf{u}}\times\hat{\mathbf{v}}.$ The intersection of these two planes is the tangent line to the circle at point $A.$ Its direction vector is the cross product $\hat{\mathbf{T}}=\hat{\mathbf{n}}\times\hat{\mathbf{v}}$ and its line equation is $(x,y,z)=A+t\cdot\hat{\mathbf{T}}.$

It might actually be less work to get the tangent vector by the more general method. However, the normal was really easy. Also this special method avoids knowing calculus. Even so, most people who can manipulate vectors at this complexity already know calculus.

General Method for the Tangent and Normal

The proper way to do this, which is a bit more general is the following.

Form the unit tangent vector. Here $\mathbf{r^{\prime}}$ is the first derivative of the circle named $\mathbf{r}.$ The tangent vector is $\mathbf{r^{\prime}},$ but to continue on the normal, we need the unit tangent vector. \[ \mathbf{T}=\frac{\mathbf{r^{\prime}}}{\mathbf{|r^{\prime}|}}=\frac{-R\cdot\sin\theta\cdot\left(\begin{array}{c} u_{x}\\ u_{y}\\ u_{z} \end{array}\right)+R\cdot\cos\left(\begin{array}{c} v_{x}\\ v_{y}\\ v_{z} \end{array}\right)}{\left\Vert -R\cdot\sin\theta\cdot\left(\begin{array}{c} u_{x}\\ u_{y}\\ u_{z} \end{array}\right)+R\cdot\cos\left(\begin{array}{c} v_{x}\\ v_{y}\\ v_{z} \end{array}\right)\right\Vert } \]

Form another unit vector from the derivative of $\mathbf{T.}$ Again, we don't need the unit normal vector and could stop with the numerator only. \[ \mathbf{N}=\frac{\mathbf{T}(t)^{\prime}}{\left|\mathbf{T}(t)^{\prime}\right|} \]

Although the notation here is not difficult, the implementation is a bit much depending on the complexity of the function. For the circle, it isn't too bad to do with numbers, but even that gets tedious with algebraic symbols, which is why it isn't shown here.