# 1.2: Distance Between Two Points; Circles

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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.