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14.4.1: Circles

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    67364
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    A circle

    A circle can be determined by fixing a point at the center and a positive number, the radius.

    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 \(r\).

    In this picture, we see that a point (h,k) is in the center of the circle and the point (x,y) is on the circle with a distance r. We express this relationship algebraically using the Distance Formula herein.

    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.

    The 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

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

    or

    $$\dfrac{(x-h)^2}{r^2} + \dfrac{(y-k)}{r^2} = 1$$

    The second form of the equation is for comparison to hyperbolas and ellipses coming in later sections.

    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

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

    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

    This is an graph of the circle defined by the equation in this problem. The circle is centered at x=-2 and y=1 with a radius of 2.

    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 \(r = \frac{5}{3}\).

    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. The reader is encouraged to think about what, if any, points lie on the graphs of these two equations.

    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

    Image of graph of circle defined by the problem herein. The center of the circle is defined as the mid-point of the two points of the circle.

    Since the given points are endpoints of a diameter, we know their midpoint \((h, k)\) is the center of the circle. This is easily solved as

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

    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 \(\left(x - \dfrac{1}{2} \right)^2 + \left(y - \dfrac{7}{2} \right)^2 =\dfrac{10}{4}\)

    The Unit Circle

    We close this section with an important circle in mathematics1: 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 \(\left(-\dfrac{1}{2}, \dfrac{\sqrt{3}}{2} \right)\).

    Contributors and Attributions

    Template:ContribZeagerPreCalc


    1The unit circle is used in trigonometry to define and evaluate trigonometric functions like sine and cosine.


    14.4.1: Circles is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by LibreTexts.