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Mathematics LibreTexts

7.2: Circles

[ "article:topic", "unit circle", "Circles", "authorname:stitzzeager", "showtoc:no" ]
  • Page ID
    4769
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    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 \(\PageIndex{1}\): Circles

    A circle with center \((h,k)\) and radius \(r>0\) 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 (Equation \ref{distanceformula}), as

    \[r = \sqrt{(x - h)^2 + (y-k)^2} \label{distanceformula}\]

    By squaring both sides of this equation, we get an equivalent equation (since \(r > 0\)) which gives us the standard equation of a circle.

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

    The equation of a circle with center \((h,k)\) and radius \(r >0\) is

    \[(x-h)^2 + (y-k)^2 = r^2. \label{standardcircle}\]

    Example \(\PageIndex{1}\): The Standard Equation

    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

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

    \(\Box\)

    Example \(\PageIndex{2}\):

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

    Solution

    From the standard form of a circle, Equation \ref{standardcircle}, 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

    \(\Box\)

    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 Equation \ref{standardcircle} by following the steps given below.

    To Write the Equation of a Circle in Standard Form

    1. Group the same variables together on one side of the equation and position the constant on the other side.
    2. Complete the square on both variables as needed.
    3. Divide both sides by the coefficient of the squares. (For circles, they will be the same.)

    Example \(\PageIndex{3}\):

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

    Solution

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

    From Equation \ref{standardcircle}, 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 \(r = \frac{5}{3}\).

    \(\Box\)

    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, Equation \ref{midpointformula}, 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. Equation \ref{midpointformula} gives us

    \[ \begin{array}{rcl} (h,k) & = & \left( \dfrac{x_{\mbox{1}} + x_{\mbox{2}}}{2}, \dfrac{y_{\mbox{1}} + y_{\mbox{2}}}{2} \right) \\ & = & \left( \dfrac{-1+2}{2}, \dfrac{3+4}{2} \right) \\ & = & \left( \dfrac{1}{2}, \dfrac{7}{2} \right) \end{array} \]

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

    \[ \begin{array}{rcl} r & = & \dfrac{1}{2} \sqrt{\left(x_{\mbox{2}} - x_{\mbox{1}}\right)^2+\left(y_{\mbox{2}}-y_{\mbox{1}}\right)^2} \\ & = & \dfrac{1}{2} \sqrt{(2-(-1))^2+(4-3)^2} \\ & = & \dfrac{1}{2} \sqrt{3^2+1^2} \\ & = &\dfrac{\sqrt{10}}{2} \end{array} \]

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

    \(\Box\)

    We close this section with the most important\footnote{While this may seem like an opinion, it is indeed a fact. See Chapters \ref{IntroTrig} and \ref{AppExt} for details.} circle in all of mathematics: the Unit Circle.

    Definition \(\PageIndex{2}\): 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 \(\dfrac{\sqrt{3}}{2}\).

    Solution

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

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

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

    \(\Box\)

    Contributors

    • Carl Stitz, Ph.D. (Lakeland Community College) and Jeff Zeager, Ph.D. (Lorain County Community College)