1.5.9: Trigonometry Preview - Circles
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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.
Circle
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, as
\[r = \sqrt{(x - h)^2 + (y-k)^2} \nonumber\]
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 Standard Form of a Circle
The standard form of the equation of a circle with center \((h,k)\) and radius \(r >0\) is
\[(x-h)^2 + (y-k)^2 = r^2. \nonumber\]
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
\[\begin{array}{rcl} (x-(-2))^2+(y-3)^2 &= &(5)^2 \\ (x+2)^2+(y-3)^2 & = & 25 \end{array} \nonumber\]
Example \(\PageIndex{2}\)
Graph \((x+2)^2+(y-1)^2 = 4\). Find the center and radius.
Solution
From the standard form of a circle, 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 to rewrite the equation in standard form for a circle.
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} \nonumber\]
From the standard form 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.
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
\[ \begin{array}{rcl} (h,k) & = & \left( \dfrac{x_{\mbox{1}} + x_{\mbox{2}}}{2}, \dfrac{y_{\mbox{1}} + y_{\mbox{2}}}{2} \right) \\ & = & \left( \dfrac{-1+2}{2}, \dfrac{3+4}{2} \right) \\ & = & \left( \dfrac{1}{2}, \dfrac{7}{2} \right) \end{array} \nonumber\]
The diameter of the circle is the distance between the given points, so we know that half of the distance is the radius. Thus,
\[ \begin{array}{rcl} r & = & \dfrac{1}{2} \sqrt{\left(x_{\mbox{2}} - x_{\mbox{1}}\right)^2+\left(y_{\mbox{2}}-y_{\mbox{1}}\right)^2} \\ & = & \dfrac{1}{2} \sqrt{(2-(-1))^2+(4-3)^2} \\ & = & \dfrac{1}{2} \sqrt{3^2+1^2} \\ & = &\dfrac{\sqrt{10}}{2} \end{array} \nonumber\]
Finally, since \(\left( \dfrac{\sqrt{10}}{2} \right)^2 = \dfrac{10}{4}\), our answer becomes \(\left(x - \dfrac{1}{2} \right)^2 + \left(y - \dfrac{7}{2} \right)^2 =\dfrac{10}{4}\)
We close this section with a very important circle that you will learn more about in trigonometry: the Unit Circle .
The 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
\[ \begin{array}{rcl} x^2 + y^2 & = & 1 \\ x^2 + \left(\dfrac{\sqrt{3}}{2}\right)^2 & = & 1 \\ \dfrac{3}{4} + x^2 & = & 1 \\ x^2 & = & \dfrac{1}{4} \\ x & = & \pm \sqrt{\dfrac{1}{4}} \\ x & = & \pm \dfrac{1}{2} \end{array} \nonumber\]
Our final answers are \(\left(\dfrac{1}{2}, \dfrac{\sqrt{3}}{2} \right)\) and \(\left(-\dfrac{1}{2}, \dfrac{\sqrt{3}}{2} \right)\).