# Drawing circles with different distance functions

1. Jan 29, 2014

### Lee33

1. The problem statement, all variables and given/known data

In $\mathbb{R}^2$, draw a unit circle for taxicab distance $(d_t)$, euclidean distance $(d_e)$, and max distance $(d_s)$.

2. Relevant equations

$d_e = \sqrt{(x -x_1)-(y-y_1)}$
$d_s = \text{max}\{|x-x_1|,|y-y_1|\}$
$d_t=|x-x_1|+|y-y_1|$

3. The attempt at a solution

Euclidean distance is the most familiar one. It will just be our everyday unit circle.

For taxicab, will the unit circle be a shape of a square?

For max distance I am not so sure about to.

2. Jan 29, 2014

### HallsofIvy

Staff Emeritus
Try a simple example like $(x_1, y_1)= (0, 0)$ and the distance is 1.

Then with the Euclidean distance you have $\sqrt{x^2+ y^2}= 1$ which is the same as $x^2+ y^2= 1$. Yes, that is a circle with center at (0, 0) and radius 1..

With the taxicab metric, that is |x|+ |y|= 1. If x and y are both positive it is x+ y= 1 which is the line segment from (1, 0) to (0, 1). If x is negative and y is positive, it is -x+ y= 1 which is the line segment from (0, 1) to (-1, 0). If x and y are both negative, it is -x- y= 1 which is the line segment from (-1, 0) to (0, -1). Finally, if x is positive and y is negative, it is x- y= 1 which is the line segment from (0, -1) to (1, 0). Yes, that is a square with its diagonals horizontal and vertical.

With the max distance it is max(|x|,|y|)= 1. If x and y are both positive and x< y that is |y|= y= 1 which is the horizontal line segment from (0, 1) to (1, 1). If x and y are both positive and y< x, that is |x|= x= 1 which is that vertical line segment from (1, 0) to (1, 1). You check the other quadrants:

If x is negative and y is positive and x< y .... If y< x ....

3. Jan 29, 2014

### Lee33

Thanks for the help!