1.3: Distance Between Two Points; Circles

Given two points $(x_1,y_1)$ and $(x_2,y_2)$, recall that their horizontal distance from one another is $\Delta x=x_2-x_1$ and their vertical distance from one another is $\Delta y=y_2-y_1$. (Actually, the word "distance'' normally denotes "positive distance''. $\Delta x$ and $\Delta y$ are signed distances, but this is clear from context.) The actual (positive) distance from one point to the other is the length of the hypotenuse of a right triangle with legs $|\Delta x|$ and $|\Delta y|$, as shown in Figure $\PageIndex{1}$. The Pythagorean theorem then says that the distance between the two points is the square root of the sum of the squares of the horizontal and vertical sides:

$$\hbox{distance} =\sqrt{(\Delta x)^2+(\Delta y)^2}=\sqrt{(x_2-x_1)^2+ (y_2-y_1)^2}. \nonumber$$

For example, the distance between points $A(2,1)$ and $B(3,3)$ is

$$\sqrt{(3-2)^2+(3-1)^2}=\sqrt{5}. \nonumber$$

Figure $\PageIndex{1}$: Distance between two points, $\Delta x$ and $\Delta y$ positive

As a special case of the distance formula, suppose we want to know the distance of a point $(x,y)$ to the origin. According to the distance formula, this is $$\sqrt{(x-0)^2+(y-0)^2}=\sqrt{x^2+y^2}. \nonumber$$

A point $(x,y)$ is at a distance $r$ from the origin if and only if

$$\sqrt{x^2+y^2}=r, \nonumber$$

or, if we square both sides:

$$x^2+y^2=r^2. \nonumber$$

This is the equation of the circle of radius $r$ centered at the origin. The special case $r=1$ is called the unit circle; its equation is

$$x^2+y^2=1. \nonumber$$

Similarly, if $C(h,k)$ is any fixed point, then a point $(x,y)$ is at a distance $r$ from the point $C$ if and only if

$$\sqrt{(x-h)^2+(y-k)^2}=r, \nonumber$$

i.e., if and only if

$$(x-h)^2+(y-k)^2=r^2. \nonumber$$

This is the equation of the circle of radius $r$ centered at the point $(h,k)$. For example, the circle of radius 5 centered at the point $(0,-6)$ has equation $(x-0)^2+(y--6)^2=25$, or $x^2+(y+6)^2=25$. If we expand this we get $x^2+y^2+12y+36=25$ or $x^2+y^2+12y+11=0$, but the original form is usually more useful.

Example $\PageIndex{1}$

Graph the circle $x^2-2x+y^2+4y-11=0$.

Solution

With a little thought we convert this to

$$(x-1)^2+(y+2)^2-16=0 \nonumber$$

or

$$(x-1)^2+(y+2)^2=16. \nonumber$$

Now we see that this is the circle with radius 4 and center $(1,-2)$, which is easy to graph.

Contributors

David Guichard (Whitman College)

• Integrated by Justin Marshall.