# What is the equation of the normal to this line?

I am playing around with the desmos graphing calculator and I am trying to find the equation of the normal to this line where it touches the curve $x^2 + y^2 = c^2$.

$$y\sin({t})+x\cos({t}) = c$$

where $c$ is the radius of the circle and $t$ is the angle in radians around the circle.

Here is a desmos graph I have created with the equation

I know that the normal line will have a negative reciprocal of the gradient of the tangent like in a very simple example

Example tangent $$y = 2x$$ Normal of example tangent $$y = -{1 \over 2}x$$

I've tried applying this to the first equation but have had no luck. I don't think I am using the correct approach...

• The normal passing through which point? – Bernard Sep 9 '18 at 0:48
• The point where the line touches the circle $x^2 + y^2 = c^2$. I'm not very good at explaining myself in a mathematical way sorry.. – Danoram Sep 9 '18 at 0:58
• Well, the equation in your post is the equation of the tangent to the circle at this point, hence the normal is just the line through this point and the centre of the circle, and its equation is simply $y=(\tan t) x$ (if $t\ne \pm\frac\pi 2$). – Bernard Sep 9 '18 at 1:10
• Yup that is what I was looking for! As easy as it sounds I have learnt more because of it – Danoram Sep 9 '18 at 1:12

## 4 Answers

Well, you can rewrite the equation $y\sin(t)+x\cos(t)=c$ as $$y=-\frac{\cos(t)}{\sin(t)}x+c \ \ \ \text{for}\ t\ne k\pi, \ k\in\mathbb{Z}.$$ The slope of this line is $m_{t}=-\frac{\cos(t)}{\sin(t)}$ so the equation of normal line that passes through the point $(x_0,y_0)$ must be $$y=-\frac{1}{m_{t}}(x-x_0)+y_0\implies y=\tan(t)(x-x_0)+y_0 \ \ \ \text{with} \ t\ne\frac{k\pi}{2}.$$ The cases $t=\frac{k\pi}{2}$ are easy.

If you want that the normal line goes through the intersection point, then you can choose $(x_0,y_0)=(0,0)$, because all the normal lines with this proprerty must contain the radius (hence $(0,0)$ must be a point of normal.) In conclusion: $y=\tan(t)x$ is the equation you are looking for.

• Thank you! of all the answers so far for me this was the easiest to understand without involving differentials etc – Danoram Sep 9 '18 at 1:11
• @Danoram You are welcome. :) – Ixion Sep 9 '18 at 1:13
• disclaimer: I appreciated all other answers too!! – Danoram Sep 9 '18 at 1:15

We can write this as $$c = y \sin(t) + x \cos(t) = (\cos(t), \sin(t)) \cdot (x, y) = n \cdot u \iff \\ n \cdot u = c$$ where $$n=(\cos(t), \sin(t))$$ is a normal vector of the line and $$u = (x, y)$$ some vector from the origin to a point on the line.

In this case $n$ is a unit normal vector which makes $\lvert c\rvert$ the distance of the line to the origin.

We see that $u=cn$ fulfills the equation of the line.

The vector $$\hat{n} = (-\sin(t), \cos(t))$$ is orthogonal to $n$, as $$\hat{n} \cdot n = (-\sin(t), \cos(t)) \cdot (\cos(t), \sin(t)) = - \sin(t)\cos(t)+\cos(t)\sin(t) = 0$$ So an equation of our normal line is $$\hat{n} \cdot u = d$$ where $\lvert d \rvert$ is the distance of the normal line to the origin, as $\hat{n}$ is a unit normal vector as well.

The point where the line touches the circle $x^2+y^2=c^2$

This is the point $u = cn$. It fulfills $$n \cdot u = n\cdot(cn) = c (n \cdot n) = c \\ (c \cos(t))^2 + (c \sin(t))^2 = c^2 (\cos(t)^2 + \sin(t)^2) = c^2$$ thus lies on the line and the circle.

We insert this $u$ in the equation of the normal line $$d = \hat{n} \cdot u = \hat{n} \cdot (cn) = c (\hat{n} \cdot n) = c \, 0 = 0$$ so we have $d=0$ and the equation $$\hat{n} \cdot u = 0$$ Note: The advantage of these normal forms is that they also work for the cases $t = \pm\pi/2$.

Starting with $x^2 + y^2 = c^2$ we can find the equation of the tangent line at the point (x,y) on the graph by implicitly differentiating both sides with respect to x.

$\frac{d}{dx} (x^2 + y^2 )=\frac{d}{dx} c^2$

$2x + 2y\frac{dy}{dx} = 0$

$\frac{dy}{dx} = -\frac{x}{y}$

To find the slope of the normal line, you can take the negative reciprocal of this value. If you want to find the slope of the normal line to begin with, just implicitly differentiate with respect to y instead. This will give you $\frac{dx}{dy}$, which is the slope of the normal line at that point.

Here's a solution with physics flavor:

You want to find the equation of the line that is normal to the circle of radius $c$ centered at origin at any time. We know the centripetal force has this property and it is in the opposite direction of the position vector. So, if $\vec{r}(t)=(c\cos(t),c\sin(t))$ is the position vector (i.e. the parametrization of the circle), the equation of the line is given by:

$$(x,y)= -c(\cos(t),\sin(t))t+c(\cos(t),\sin(t))$$ $$x=c\cos(t)(1-t) \hspace{10px}(1)$$ $$y=c\sin(t)(1-t) \hspace{10px}(2)$$

Assuming that $\cos(t)\neq 0$, solve for $t$ in$(1)$ and plug it in $(2)$:

$$t=(1-\frac{x}{c\cos(t)})$$

$$y=c\sin(t)\big(1-(1-\frac{x}{c\cos(t)})\big)=\tan(t)x$$