$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$

# 1.2: Distance Between Two Points; Circles

• • Contributed by David Guichard
• Professor (Mathematics) at Whitman College

$$\newcommand{\vecs}{\overset { \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

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:

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

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

$$\sqrt{(3-2)^2+(3-1)^2}=\sqrt{5}.$$ 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 $$\sqrt{(x-0)^2+(y-0)^2}=\sqrt{x^2+y^2}.$$

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

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

or, if we square both sides:

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

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

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

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

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

i.e., if and only if

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

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

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

or

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

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

## Contributors

• Integrated by Justin Marshall.