To be honest, it should be called Polar Coordinate System

In mathematics, the polar coordinate system is a two-dimensional coordinate system in which each point on a plane is determined by a distance from a reference point and an angle from a reference direction.

• The reference point (analogous to the origin of a Cartesian coordinate system, the center point) is called the pole
• The ray from the pole in the reference direction is the polar axis

You can represent any point in this system with (distance, angle), for example, (3, 60 degree)

or generally, $(\rho, \theta)$

$\rho$: rho

$\theta$: theta

Converting between polar and Cartesian coordinates

$$x = \rho \cdot \cos{\theta}$$

$$y = \rho \cdot \sin{\theta}$$

$$\rho = \sqrt{x^2 + y^2}$$

Question and Answers

In polar coordinate system, a circle equation is $\rho = 8 \cdot \sin{\theta}$, a line equation is $\theta=\frac{\pi}{3}$, what's the max distance for a point at that circle to the line?

the answer is 6

First, we need to convert all information from polar coordinate to Cartesian coordinate:

1. circle

$$\begin{align*} \rho &= 8 \cdot \sin{\theta} \\ \rho &= 8 \cdot \frac{y}{\rho} \\ {\rho}^2 &= 8y \\ x^2 + y^2 &= 8y \\ x^2 + y^2 - 8y &= 0 \\ x^2 + (y-4)^2 &= 16 = 4^2 \\ \end{align*}$$ so now we know the center point of that circle is (0, 4), radius of the circle is 4

2. line

we know $\theta=\frac{\pi}{3}$ is nothing but a line of 60 degree angle.

And we also know a general line could be represented as $y = slope \cdot x$

And $slope = \tan(\theta)$

so: $$\begin{align*} slope &= \tan\frac{\pi}{3} \\ &= \sqrt{3} \\ \\ y &= \sqrt{3} \cdot x \\ \end{align*}$$

How to get the max distance?

Inner a circle, diameter is the longest line segment you can get.

A special line drawn inside a circle is called a chord. A chord can be of different lengths. The longest length of the chord is called a diameter. Two times the radius makes a diameter.

And for the distance between a point($x_0, y_0$) and a line($ax + by + c = 0$), there's a formula for it:

$$distance = \frac{|a(x_0) + b(y_0) + c|}{\sqrt{a^2 + b^2}}$$

So first, we need to choose a line as close a diameter as possible.

For some reason, people already known the distance between a line and circle center + the radius of that circle = the max distance for a point at that circle to the line.

so: $$\begin{align*} \text{distance between circle center and line} &= \frac{|\sqrt{3} \cdot 0 + (-1) \cdot (-4)|}{\sqrt{\sqrt{3}^2 + 1^2}} \\ &= \frac{4}{2} \\ &= 2 \\ \\ \text{max distance} &= 4 + \text{radius of that circle} \\ \text{max distance} &= 4 + 2 \\ \text{max distance} &= 6 \end{align*}$$

Thank you for reading, I love the feeling of doing this.