This is “Circles”, section 8.2 from the book Advanced Algebra (v. 1.0). For details on it (including licensing), click here.
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A circleA circle is the set of points in a plane that lie a fixed distance from a given point, called the center. is the set of points in a plane that lie a fixed distance, called the radiusThe fixed distance from the center of a circle to any point on the circle., from any point, called the center. The diameterThe length of a line segment passing through the center of a circle whose endpoints are on the circle. is the length of a line segment passing through the center whose endpoints are on the circle. In addition, a circle can be formed by the intersection of a cone and a plane that is perpendicular to the axis of the cone:
In a rectangular coordinate plane, where the center of a circle with radius r is $\left(h,k\right)$, we have
Calculate the distance between $\left(h,k\right)$ and $\left(x,y\right)$ using the distance formula,
$$\sqrt{{\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}}=r$$
Squaring both sides leads us to the equation of a circle in standard formThe equation of a circle written in the form ${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={r}^{2}$ where $\left(h,k\right)$ is the center and r is the radius.,
$${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={r}^{2}$$
In this form, the center and radius are apparent. For example, given the equation ${\left(x-2\right)}^{2}+\text{\hspace{0.17em}}{\left(y\text{\hspace{0.17em}}+\text{\hspace{0.17em}}5\right)}^{2}=16$ we have,
$$\begin{array}{ccccc}{\left(x-h\right)}^{2}& +& \hspace{0.17em}{\left(x-k\right)}^{2}& =& {r}^{2}\hfill \\ \phantom{\rule{1.0em}{0ex}}{\downarrow}& & \phantom{\rule{1.5em}{0ex}}{\downarrow}& & {\downarrow}\phantom{\rule{1.5em}{0ex}}\\ {\left(x-{2}\right)}^{2}& +& {\left[y-\left({-5}\right)\right]}^{2}& =& {{4}}^{2}\hfill \end{array}$$
In this case, the center is $\left(2,-5\right)$ and $r=4.$ More examples follow:
Equation |
Center |
Radius |
---|---|---|
${\left(x-3\right)}^{2}+{\left(y-4\right)}^{2}=25$ |
$\left(3,4\right)$ |
$r=5$ |
${\left(x-1\right)}^{2}+{\left(y+2\right)}^{2}=7$ |
$\left(1,-2\right)$ |
$r=\sqrt{7}$ |
${\left(x+4\right)}^{2}+{\left(y-3\right)}^{2}=1$ |
$\left(-4,3\right)$ |
$r=1$ |
${x}^{2}+{\left(y+6\right)}^{2}=8$ |
$\left(0,-6\right)$ |
$r=2\sqrt{2}$ |
The graph of a circle is completely determined by its center and radius.
Graph: ${\left(x-2\right)}^{2}+{\left(y+5\right)}^{2}=16.$
Solution:
Written in this form we can see that the center is $\left(2,-5\right)$ and that the radius $r=4$ units. From the center mark points 4 units up and down as well as 4 units left and right.
Then draw in the circle through these four points.
Answer:
As with any graph, we are interested in finding the x- and y-intercepts.
Find the intercepts: ${\left(x-2\right)}^{2}+{\left(y+5\right)}^{2}=16.$
Solution:
To find the y-intercepts set $x=0$:
$$\begin{array}{ccc}\hfill {\left(x-2\right)}^{2}+{\left(y+5\right)}^{2}& =& 16\hfill \\ \hfill {\left({0}-2\right)}^{2}+{\left(y+5\right)}^{2}& =& 16\hfill \\ \hfill 4+{\left(y+5\right)}^{2}& =& 16\hfill \end{array}$$
For this equation, we can solve by extracting square roots.
$$\begin{array}{ccc}\hfill {\left(y+5\right)}^{2}& =& 12\hfill \\ \hfill y+5& =& \pm \sqrt{12}\hfill \\ \hfill y+5& =& \pm 2\sqrt{3}\hfill \\ \hfill y& =& -5\pm 2\sqrt{3}\hfill \end{array}$$
Therefore, the y-intercepts are $\left(0,-5-2\sqrt{3}\right)$ and $\left(0,-5+2\sqrt{3}\right).$ To find the x-intercepts set $y=0$:
$$\begin{array}{ccc}\hfill {\left(x-2\right)}^{2}+{\left(y+5\right)}^{2}& =& 16\hfill \\ \hfill {\left(x-2\right)}^{2}+{\left({0}+5\right)}^{2}& =& 16\hfill \\ \hfill {\left(x-2\right)}^{2}+25& =& 16\hfill \\ \hfill {\left(x-2\right)}^{2}& =& -9\hfill \\ \hfill x-2& =& \pm \sqrt{-9}\hfill \\ \hfill x& =& 2\pm 3i\hfill \end{array}$$
And because the solutions are complex we conclude that there are no real x-intercepts. Note that this does make sense given the graph.
Answer: x-intercepts: none; y-intercepts: $\left(0,-5-2\sqrt{3}\right)$ and $\left(0,-5+2\sqrt{3}\right)$
Given the center and radius of a circle, we can find its equation.
Graph the circle with radius $r=3$ units centered at $\left(-1,0\right).$ Give its equation in standard form and determine the intercepts.
Solution:
Given that the center is $\left(-1,0\right)$ and the radius is $r=3$ we sketch the graph as follows:
Substitute h, k, and r to find the equation in standard form. Since $\left(h,k\right)=\left(-1,0\right)$ and $r=3$ we have,
$$\begin{array}{ccc}\hfill {\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}& =& {r}^{2}\hfill \\ \hfill {\left[x-\left({-}{1}\right)\right]}^{2}+{\left(y-{0}\right)}^{2}& =& {{3}}^{2}\hfill \\ \hfill {\left(x+1\right)}^{2}+{y}^{2}& =& 9\hfill \end{array}$$
The equation of the circle is ${\left(x+1\right)}^{2}+{y}^{2}=9$, use this to determine the y-intercepts.
$$\begin{array}{cccc}\hfill {\left(x+1\right)}^{2}+{y}^{2}& =& 9\hfill & \text{\hspace{0.17em}}\text{\hspace{0.17em}}{Set\text{\hspace{0.17em}}x=0\text{\hspace{0.17em}}to\text{\hspace{0.17em}}and\text{\hspace{0.17em}}solve\text{\hspace{0.17em}}for\text{\hspace{0.17em}}y.}\\ \hfill {\left({0}+1\right)}^{2}+{y}^{2}& =& 9\hfill & \\ \hfill 1+{y}^{2}& =& 9\hfill & \\ \hfill {y}^{2}& =& 8\hfill & \\ \hfill y& =& \pm \sqrt{8}\hfill & \\ \hfill y& =& \pm 2\sqrt{2}\hfill & \end{array}$$
Therefore, the y-intercepts are $\left(0,-2\sqrt{2}\right)$ and $\left(0,2\sqrt{2}\right).$ To find the x-intercepts algebraically, set $y=0$ and solve for x; this is left for the reader as an exercise.
Answer: Equation: ${\left(x+1\right)}^{2}+{y}^{2}=9$; y-intercepts: $\left(0,-2\sqrt{2}\right)$ and $\left(0,2\sqrt{2}\right)$; x-intercepts: $\left(-4,0\right)$ and $\left(2,0\right)$
Of particular importance is the unit circleThe circle centered at the origin with radius 1; its equation is ${x}^{2}+{y}^{2}=1.$,
$${x}^{2}+{y}^{2}=1$$
Or,
$${\left(x-0\right)}^{2}+{\left(y-0\right)}^{2}={1}^{2}$$
In this form, it should be clear that the center is $\left(0,0\right)$ and that the radius is 1 unit. Furthermore, if we solve for y we obtain two functions:
$$\begin{array}{ccc}\hfill {x}^{2}+{y}^{2}& =& 1\hfill \\ \hfill {y}^{2}& =& 1-{x}^{2}\hfill \\ \hfill y& =& \pm \sqrt{1-{x}^{2}}\hfill \end{array}$$
The function defined by $y=\sqrt{1-{x}^{2}}$ is the top half of the circle and the function defined by $y=-\sqrt{1-{x}^{2}}$ is the bottom half of the unit circle:
Try this! Graph and label the intercepts: ${x}^{2}+{\left(y+2\right)}^{2}=25.$
Answer:
We have seen that the graph of a circle is completely determined by the center and radius which can be read from its equation in standard form. However, the equation is not always given in standard form. The equation of a circle in general formThe equation of a circle written in the form ${x}^{2}+{y}^{2}+cx+dy+e=0.$ follows:
$${x}^{2}+{y}^{2}+cx+dy+e=0$$
Here c, d, and e are real numbers. The steps for graphing a circle given its equation in general form follow.
Graph: ${x}^{2}+{y}^{2}+6x-8y+13=0.$
Solution:
Begin by rewriting the equation in standard form.
Step 1: Group the terms with the same variables and move the constant to the right side. In this case, subtract 13 on both sides and group the terms involving x and the terms involving y as follows.
$$\begin{array}{ccc}\hfill {x}^{2}+{y}^{2}+6x-8y+13& =& 0\hfill \\ \hfill \left({x}^{2}+6x+\_\_\_\right)+\left({y}^{2}-8y+\_\_\_\right)& =& -13\hfill \end{array}$$
Step 2: Complete the square for each grouping. The idea is to add the value that completes the square, ${\left(\frac{b}{2}\right)}^{2}$, to both sides for both groupings, and then factor. For the terms involving x use ${\left(\frac{6}{2}\right)}^{2}={3}^{2}=9$ and for the terms involving y use ${\left(\frac{-8}{2}\right)}^{2}={\left(-4\right)}^{2}=16.$
$$\begin{array}{ccc}\hfill \left({x}^{2}+6x{\text{\hspace{0.17em}}+}{9}\right)+\left({y}^{2}-8y{+}{16}\right)& =& -13{\text{\hspace{0.17em}}+}{9}{+}{16}\hfill \\ \hfill {\left(x+3\right)}^{2}+{\left(y-4\right)}^{2}& =& 12\hfill \end{array}$$
Answer:
Determine the center and radius: $4{x}^{2}+4{y}^{2}-8x+12y-3=0.$
Solution:
We can obtain the general form by first dividing both sides by 4.
$$\begin{array}{ccc}\hfill \frac{4{x}^{2}+4{y}^{2}-8x+12y-3}{4}& =& \frac{0}{4}\hfill \\ \hfill {x}^{2}+{y}^{2}-2x+3y-\frac{3}{4}& =& 0\hfill \end{array}$$
Now that we have the general form for a circle, where both terms of degree two have a leading coefficient of 1, we can use the steps for rewriting it in standard form. Begin by adding $\frac{3}{4}$ to both sides and group variables that are the same.
$$\left({x}^{2}-2x+\_\_\_\right)+\left({y}^{2}+3y+\_\_\_\right)=\frac{3}{4}$$
Next complete the square for both groupings. Use ${\left(\frac{-2}{2}\right)}^{2}={\left(-1\right)}^{2}=1$ for the first grouping and ${\left(\frac{3}{2}\right)}^{2}=\frac{9}{4}$ for the second grouping.
$$\begin{array}{ccc}\hfill \left({x}^{2}-2x{\text{\hspace{0.17em}}+}{1}\right)+\left({y}^{2}+3y{+}{\frac{9}{4}}\right)& =& \frac{3}{4}{\text{\hspace{0.17em}}+}{1}{+}{\frac{9}{4}}\hfill \\ \hfill {\left(x-1\right)}^{2}+{\left(y+\frac{3}{2}\right)}^{2}& =& \frac{16}{4}\hfill \\ \hfill {\left(x-1\right)}^{2}+{\left(y+\frac{3}{2}\right)}^{2}& =& 4\hfill \end{array}$$
Answer: Center: $\left(1,-\frac{3}{2}\right)$; radius: $r=2$
In summary, to convert from standard form to general form we multiply, and to convert from general form to standard form we complete the square.
Determine the center and radius given the equation of a circle in standard form.
${\left(x-5\right)}^{2}+{\left(y+4\right)}^{2}=64$
${\left(x+9\right)}^{2}+{\left(y-7\right)}^{2}=121$
${x}^{2}+{\left(y+6\right)}^{2}=4$
${\left(x-1\right)}^{2}+{y}^{2}=1$
${\left(x+1\right)}^{2}+{\left(y+1\right)}^{2}=7$
${\left(x+2\right)}^{2}+{\left(y-7\right)}^{2}=8$
Determine the standard form for the equation of the circle given its center and radius.
Center $\left(5,7\right)$ with radius $r=7.$
Center $\left(-2,8\right)$ with radius $r=5.$
Center $\left(6,-11\right)$ with radius $r=\sqrt{2}.$
Center $\left(-4,-5\right)$ with radius $r=\sqrt{6}.$
Center $\left(0,-1\right)$ with radius $r=2\sqrt{5}.$
Center $\left(0,0\right)$ with radius $r=3\sqrt{10}.$
Graph.
${\left(x-1\right)}^{2}+{\left(y-2\right)}^{2}=9$
${\left(x+3\right)}^{2}+{\left(y-3\right)}^{2}=25$
${\left(x-2\right)}^{2}+{\left(y+6\right)}^{2}=4$
${\left(x+6\right)}^{2}+{\left(y+4\right)}^{2}=36$
${x}^{2}+{\left(y-4\right)}^{2}=1$
${\left(x-3\right)}^{2}+{y}^{2}=4$
${x}^{2}+{y}^{2}=12$
${x}^{2}+{y}^{2}=8$
${\left(x-7\right)}^{2}+{\left(y-6\right)}^{2}=2$
${\left(x+2\right)}^{2}+{\left(y-5\right)}^{2}=5$
${\left(x+3\right)}^{2}+{\left(y-1\right)}^{2}=18$
${\left(x-3\right)}^{2}+{\left(y-2\right)}^{2}=15$
Find the x- and y-intercepts.
${\left(x-1\right)}^{2}+{\left(y-2\right)}^{2}=9$
${\left(x+5\right)}^{2}+{\left(y-3\right)}^{2}=25$
${x}^{2}+{\left(y-4\right)}^{2}=1$
${\left(x-3\right)}^{2}+{y}^{2}=18$
${x}^{2}+{y}^{2}=50$
${x}^{2}+{\left(y+9\right)}^{2}=20$
${\left(x-4\right)}^{2}+{\left(y+5\right)}^{2}=10$
${\left(x+10\right)}^{2}+{\left(y-20\right)}^{2}=400$
Find the equation of the circle.
Circle with center $\left(1,-2\right)$ passing through $\left(3,-4\right).$
Circle with center $\left(-4,-1\right)$ passing through $\left(0,-3\right).$
Circle whose diameter is defined by $\left(5,1\right)$ and $\left(-1,7\right).$
Circle whose diameter is defined by $\left(-5,7\right)$ and $\left(-1,-5\right).$
Circle with center $\left(5,-2\right)$ and area $9\pi $ square units.
Circle with center $\left(-8,-3\right)$ and circumference $12\pi $ square units.
Find the area of the circle with equation ${\left(x+12\right)}^{2}+{\left(x-5\right)}^{2}=7.$
Find the circumference of the circle with equation ${\left(x+1\right)}^{2}+{\left(y+5\right)}^{2}=8.$
Rewrite in standard form and graph.
${x}^{2}+{y}^{2}+4x-2y-4=0$
${x}^{2}+{y}^{2}-10x+2y+10=0$
${x}^{2}+{y}^{2}+2x+12y+36=0$
${x}^{2}+{y}^{2}-14x-8y+40=0$
${x}^{2}+{y}^{2}+6y+5=0$
${x}^{2}+{y}^{2}-12x+20=0$
${x}^{2}+{y}^{2}+8x+12y+16=0$
${x}^{2}+{y}^{2}-20x-18y+172=0$
$4{x}^{2}+4{y}^{2}-4x+8y+1=0$
$9{x}^{2}+9{y}^{2}+18x+6y+1=0$
${x}^{2}+{y}^{2}+4x+8y+14=0$
${x}^{2}+{y}^{2}-2x-4y-15=0$
${x}^{2}+{y}^{2}-x-2y+1=0$
${x}^{2}+{y}^{2}-x+y-\frac{1}{2}=0$
$4{x}^{2}+4{y}^{2}+8x-12y+5=0$
$9{x}^{2}+9{y}^{2}+12x-36y+4=0$
$2{x}^{2}+2{y}^{2}+6x+10y+9=0$
$9{x}^{2}+9{y}^{2}-6x+12y+4=0$
Given a circle in general form, determine the intercepts.
${x}^{2}+{y}^{2}-5x+3y+6=0$
${x}^{2}+{y}^{2}+x-2y-7=0$
${x}^{2}+{y}^{2}-6y+2=2$
${x}^{2}+{y}^{2}-6x-8y+5=0$
$2{x}^{2}+2{y}^{2}-3x-9=0$
$3{x}^{2}+3{y}^{2}+8y-16=0$
Determine the area of the circle whose equation is ${x}^{2}+{y}^{2}-2x-6y-35=0.$
Determine the area of the circle whose equation is $4{x}^{2}+4{y}^{2}-12x-8y-59=0.$
Determine the circumference of a circle whose equation is ${x}^{2}+{y}^{2}-5x+1=0.$
Determine the circumference of a circle whose equation is ${x}^{2}+{y}^{2}+5x-2y+3=0.$
Find general form of the equation of a circle centered at $\left(-3,5\right)$ passing through $\left(1,-2\right).$
Find general form of the equation of a circle centered at $\left(-2,-3\right)$ passing through $\left(-1,3\right).$
Given the graph of a circle, determine its equation in general form.
Is the center of a circle part of the graph? Explain.
Make up your own circle, write it in general form, and graph it.
Explain how we can tell the difference between the equation of a parabola in general form and the equation of a circle in general form. Give an example.
Do all circles have intercepts? What are the possible numbers of intercepts? Illustrate your explanation with graphs.
Center: $\left(5,-4\right)$; radius: $r=8$
Center: $\left(0,-6\right)$; radius: $r=2$
Center: $\left(-1,-1\right)$; radius: $r=\sqrt{7}$
${\left(x-5\right)}^{2}+{\left(y-7\right)}^{2}=49$
${\left(x-6\right)}^{2}+{\left(y+11\right)}^{2}=2$
${x}^{2}+{\left(y+1\right)}^{2}=20$
x-intercepts: $\left(1\pm \sqrt{5},0\right)$; y-intercepts: $\left(0,2\pm 2\sqrt{2}\right)$
x-intercepts: none; y-intercepts: $\left(0,3\right)$, $\left(0,5\right)$
x-intercepts: $\left(\pm 5\sqrt{2},0\right)$; y-intercepts: $\left(0,\pm 5\sqrt{2}\right)$
x-intercepts: none; y-intercepts: none
${\left(x-1\right)}^{2}+{\left(y+2\right)}^{2}=8$
${\left(x-2\right)}^{2}+{\left(y-4\right)}^{2}=18$
${\left(x-5\right)}^{2}+{\left(y+2\right)}^{2}=9$
$7\pi $ square units
${\left(x+2\right)}^{2}+{\left(y-1\right)}^{2}=9$;
${\left(x+1\right)}^{2}+{\left(y+6\right)}^{2}=1$;
${x}^{2}+{\left(y+3\right)}^{2}=4$;
${\left(x+4\right)}^{2}+{\left(y+6\right)}^{2}=36$;
${\left(x-\frac{1}{2}\right)}^{2}+{\left(y+1\right)}^{2}=1$;
${\left(x+2\right)}^{2}+{\left(y-4\right)}^{2}=6$;
${\left(x-\frac{1}{2}\right)}^{2}+{\left(y-1\right)}^{2}=\frac{1}{4}$;
${\left(x+1\right)}^{2}+{\left(y-\frac{3}{2}\right)}^{2}=2$;
${\left(x+\frac{3}{2}\right)}^{2}+{\left(y+\frac{5}{2}\right)}^{2}=4$;
x-intercepts: $\left(2,0\right)$, $\left(3,0\right)$; y-intercepts: none
x-intercepts: $\left(0,0\right)$; y-intercepts: $\left(0,0\right)$, $\left(0,6\right)$
x-intercepts: $\left(-\frac{3}{2},0\right)$, $\left(3,0\right)$; y-intercepts: $\left(0,\pm \frac{3\sqrt{2}}{2}\right)$
$45\pi $ square units
$\pi \sqrt{21}$ units
${x}^{2}+{y}^{2}+6x-10y-31=0$
${x}^{2}+{y}^{2}-6x+10y+18=0$
${x}^{2}+{y}^{2}+2y=0$
Answer may vary
Answer may vary