8.2: Circles

Recall from Geometry that a circle can be determined by fixing a point (called the center) and a positive number (called the radius) as follows.

Definition: Circle

A circle with center \((h,k)\) and radius \(r\gt 0\) is the set of all points \((x, y)\) in the plane whose distance to \((h,k)\) is \(r\).

From the picture, we see that a point \((x,y)\) is on the circle if and only if its distance to \((h,k)\) is \(r\). We express this relationship algebraically using the Distance Formula as \[r = \sqrt{(x - h)^2 + (y-k)^2}\nonumber\]By squaring both sides of this equation, we get an equivalent equation (since \(r \gt 0\)) which gives us the standard equation of a circle.

Theorem \(\PageIndex{1}\): Standard Equation of a Circle

The standard equation of a circle with center \((h,k)\) and radius \(r \gt 0\) is \((x-h)^2 + (y-k)^2 = r^2.\)

Example \( \PageIndex{1} \)

Write the standard equation of the circle with center \((-2,3)\) and radius \(5\).

Solution

Here, \((h,k) = (-2,3)\) and \(r = 5\), so we get

\[\begin{array}{rrcl}
& (x-(-2))^2+(y-3)^2 & = & (5)^2 \\
\implies & (x+2)^2+(y-3)^2 & = & 25 \\
\end{array}\nonumber\]

Example \( \PageIndex{2} \)

Graph \((x+2)^2+(y-1)^2 = 4\). Find the center and radius.

Solution

From the standard form of a circle we have that \(x + 2\) is \(x-h\), so \(h = -2\) and \(y - 1\) is \(y - k\) so \(k = 1\). This tells us that our center is \((-2,1)\). Furthermore, \(r^2 = 4\), so \(r = 2\). Thus we have a circle centered at \((-2,1)\) with a radius of \(2\). Graphing gives us

If we were to expand the equation in the previous example and gather up like terms, instead of the easily recognizable \((x+2)^2 + (y-1)^2 = 4\), we’d be contending with \(x^2 + 4x + y^2 - 2y + 1 = 0.\) If we’re given such an equation, we can complete the square in each of the variables to see if it fits the form given in the standard equation of a circle. The following example illustrates this process.

Example \( \PageIndex{3} \)

Complete the square to find the center and radius of \(3x^2 - 6x + 3y^2 + 4y -4 = 0\).

Solution

\[\begin{array}{rrclr}
& 3x^2 - 6x + 3y^2 + 4y -4 & = & 0 & \\
\implies & 3x^2 - 6x + 3y^2 + 4y & = & 4 & \left(\text{add 4 to both sides}\right) \\
\implies & 3\left(x^2 - 2x \right) + 3\left(y^2 + \dfrac{4}{3} y\right) & = & 4 & \left(\text{factor out leading coefficients}\right) \\
\implies & 3\left(x^2 - 2x + \underline{1} \right) + 3\left(y^2 + \dfrac{4}{3} y + \underline{\underline{\dfrac{4}{9}}} \right) & = & 4 + 3\underline{(1)} + 3\underline{\underline{\left(\dfrac{4}{9}\right)}} & \left(\text{complete the square}\right) \\
\implies & 3(x - 1)^2 + 3\left(y + \dfrac{2}{3}\right)^2 & = & \dfrac{25}{3} & \left(\text{factor}\right) \\
\implies & (x - 1)^2 + \left(y + \dfrac{2}{3}\right)^2 & = & \dfrac{25}{9} & \left(\text{divide both sides by 3}\right) \\
\end{array}\nonumber\]

From the standard equation of a circle, we identify \(x - 1\) as \(x - h\), so \(h = 1\), and \(y + \frac{2}{3}\) as \(y - k\), so \(k = - \frac{2}{3}\). Hence, the center is \((h,k) = \left(1, -\frac{2}{3}\right)\). Furthermore, we see that \(r^2 = \frac{25}{9}\) so the radius is \(r = \frac{5}{3}\).

It is possible to obtain equations like \((x-3)^2 + (y+1)^2 = 0\) or \((x-3)^2 + (y+1)^2 = -1\), neither of which describes a circle. (Do you see why not?) The reader is encouraged to think about what, if any, points lie on the graphs of these two equations. The next example uses the Midpoint Formula in conjunction with the ideas presented so far in this section.

Example \( \PageIndex{4} \)

Write the standard equation of the circle which has \((-1,3)\) and \((2,4)\) as the endpoints of a diameter.

Solution

We recall that a diameter of a circle is a line segment containing the center and two points on the circle. Plotting the given data yields

Since the given points are endpoints of a diameter, we know their midpoint \((h, k)\) is the center of the circle. The Midpoint Formula gives us

\[\begin{array}{rcl}
(h,k) & = & \left( \dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2} \right) \\
& = & \left( \dfrac{-1+2}{2}, \dfrac{3+4}{2} \right) \\
& = & \left( \dfrac{1}{2}, \dfrac{7}{2} \right) \\
\end{array}\nonumber\]

The diameter of the circle is the distance between the given points, so we know that half of the distance is the radius. Thus,

\[\begin{array}{rcl}
r & = & \dfrac{1}{2} \sqrt{\left(x_2 - x_1\right)^2+\left(y_2-y_1\right)^2} \\
& = & \dfrac{1}{2} \sqrt{(2-(-1))^2+(4-3)^2} \\
& = & \dfrac{1}{2} \sqrt{3^2+1^2} \\
& = &\dfrac{\sqrt{10}}{2} \\
\end{array}\nonumber\]

Finally, since \(\left( \frac{\sqrt{10}}{2} \right)^2 = \frac{10}{4}\), our answer becomes \(\left(x - \frac{1}{2} \right)^2 + \left(y - \frac{7}{2} \right)^2 =\frac{10}{4}\)

We close this section with the most important¹ circle in all of mathematics: the unit circle.

Definition: Unit Circle

The unit circle is the circle centered at \((0,0)\) with a radius of \(1\). The standard equation of the Unit Circle is \(x^2 + y^2 = 1.\)

Example \( \PageIndex{5} \)

Find the points on the unit circle with \(y\)-coordinate \(\frac{\sqrt{3}}{2}\).

Solution

We replace \(y\) with \(\frac{\sqrt{3}}{2}\) in the equation \(x^2 + y^2 = 1\) to get

\[\begin{array}{rrcl}
& x^2 + y^2 & = & 1 \\
\implies &  x^2 + \left(\dfrac{\sqrt{3}}{2}\right)^2 & = & 1 \\
\implies &  \dfrac{3}{4} + x^2 & = & 1 \\
\implies &  x^2 & = & \dfrac{1}{4} \\
\implies &  x & = & \pm \sqrt{\dfrac{1}{4}} \\
\implies &  x & = & \pm \dfrac{1}{2} \\
\end{array}\nonumber\]

Our final answers are \(\left(\frac{1}{2}, \frac{\sqrt{3}}{2} \right)\) and \(\left(-\frac{1}{2}, \frac{\sqrt{3}}{2} \right)\).

Section Footnotes

¹ While this may seem like an opinion, it is indeed a fact. See Chapters 10 and 11 for details.