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College Algebra/Analytic Geometry/Circles

Analytic Geometry

Contents

Circles

Parts of a Circle

A circle in the coordinate plane can be identified by its center and radius and can be defined by an equation.

A circle is the set of all points that are a fixed distance rr from the center. The distance from the center to any point on the circle is the radius. For example, when a compass is used to draw a circle, the point is placed at the center, and the width of the compass opening is the radius. Because the width is always the same, as the compass is turned, the pencil draws a circle.

Suppose the point (h,k)(h,k) represents the center of a circle. For any point (x,y)(x,y) in the coordinate plane, the distance from (x,y)(x,y) to (h,k)(h,k) is given by the distance formula as:
(x−h)2+(y−k)2\sqrt{(x-h)^2+(y-k)^2}
The set of points that have a distance of rr from the center can be written as:
r=(x−h)2+(y−k)2r=\sqrt{(x-h)^2+(y-k)^2}
Squaring both sides of this equation produces the standard form of the equation of a circle with its center at the point (h,k)(h, k) and a radius of rr units:
(x−h)2+(y−k)2=r2(x-h)^2+(y-k)^2= r^2
If the circle is centered at the origin, then the equation can be written as:
(x−0)2+(y−0)2=r2x2+y2=r2\begin{aligned}(x-0)^2+(y-0)^2=r^2\\x^2+y^2= r^2\end{aligned}

Circles in the Coordinate Plane

Circles Centered at the Origin
x2+y2=r2x^2+y^2= r^2
 Circles Centered at (h,k)(h,k)
(x−h)2+(y−k)2=r2(x-h)^2+(y-k)^2= r^2
A: x2+y2=25x^2+y^2=25
     r=5r = 5
B: x2+y2=49x^2+y^2=49
     r=7r = 7
C: x2+y2=81x^2+y^2=81
     r=9r = 9 		D: (x−8)2+(y−8)2=36(x-8)^2 + (y-8)^2=36
     Center (8,8)(8,8) , r=6r=6
E: (x+9)2+(y−7)2=64(x+9)^2 + (y-7)^2=64
     Center (−9,7)(-9, 7) , r=8r=8
F: x2+(y+5)2=25x^2+(y+5)^2=25
     Center (0,−5)(0, -5) , r=5r=5
G: (x+10)2+(y+5)2=4(x+10)^2 + (y+5)^2=4
     Center (−10,−5)(-10, -5) , r=2r=2

Graphing Circles

A circle with a given equation can be graphed in the coordinate plane by determining the center and radius.
When the equation of a circle is not given in standard form, the center and radius may be difficult to identify. Sometimes, the equation of a circle is given in general form:
Ax2+By2+Cx+Dy+E=0Ax^2+By^2+Cx+Dy+E=0
It is called the general form because any conic section can be written in that form. The coefficients determine the type of conic section represented by the equation. If A=BA=B and A≠0A\neq0, then the graph of the equation is a circle. To identify the center and radius and to graph the circle, it is helpful to rewrite it in standard form:
(x−h)2+(y−k)2=r2(x-h)^2+(y-k)^2=r^2
The process involves completing the square, a method of rewriting parts of the equation as a perfect square.
Step-By-Step Example
Graphing a Circle by Completing the Square
Graph the circle with the equation:
x2−12x+y2+6y−19=0x^2-12x+y^2+6y-19=0
Step 1
Write the equation in standard form. Move the constant term to the right side of the equation, and group terms with the same variable.
x2−12x+y2+6y−19=0x2−12x+y2+6y=19(x2−12x)+(y2+6y)=19\begin{aligned}{\color{#c42126}x^2-12x}+{\color{#0047af}y^2+6y}-19&=0\\{\color{#c42126}x^2-12x}+{\color{#0047af}y^2+6y}&=19\\({\color{#c42126}x^2-12x})+({\color{#0047af}y^2+6y})&=19\end{aligned}
Step 2
Identify the terms that need to be added to complete the square for the xx -terms and the yy-terms. Write the equation, using blanks for the missing terms.
(x2−12x+    ‾)+(y2+6y+    ‾)=19+    ‾+    ‾\begin{aligned}({\color{#c42126} x^2-12x+\underline{\ \ \ \ }})+ ({\color{#0047af} y^2+6y+\underline{\ \ \ \ }})=19+{\color{#c42126} \underline{\ \ \ \ }}+{\color{#0047af} \underline{\ \ \ \ }}\end{aligned}
The missing term for each variable is a number that will make each expression a perfect square. To find this number, divide the coefficient of the middle term by 2, and square the result. For the xx-terms:
(−122)2=36\left(\frac{-12}{2}\right)^2=36
For the yy-terms:
(62)2=9\left(\frac{6}{2}\right)^2=9
Step 3
Complete the square for each variable. Add the missing number for the xx -terms to both sides of the equation, and add the missing number for the yy -terms to both sides of the equation. Then, factor the expressions on the left side into perfect squares and simplify the right side to write the equation in standard form.
(x2−12x+    ‾)+(y2+6y+    ‾)=19+    ‾+    ‾(x2−12x+36)+(y2+6y+9)=19+36+9(x−6)2+(y+3)2=64\begin{aligned}({\color{#c42126} x^2-12x+\underline{\ \ \ \ }})+ ({\color{#0047af} y^2+6y+\underline{\ \ \ \ }})&=19+{\color{#c42126} \underline{\ \ \ \ }}+{\color{#0047af} \underline{\ \ \ \ }}\\({\color{#c42126} x^2-12x+36})+({\color{#0047af} y^2+6y+9})&=19+{\color{#c42126} 36}+{\color{#0047af} 9}\\({\color{#c42126} x-6})^2+({\color{#0047af} y+3})^2&=64\end{aligned}
Step 4
Identify the center and radius.
(x−6)2+(y+3)2=64({\color{#c42126} x-6})^2+({\color{#0047af} y+3})^2=64
The center of the circle is the point (6,−3)({\color{#c42126}6}, {\color{#0047af}-3}), and the radius is 64=8\sqrt{64}=8.
Solution

By Hand: Plot the center of the circle. Use the radius to plot additional points that are 8 units up, down, left, and right from the center, and graph the circle through these points.

On a Graphing Calculator: Solve for yy.
(x−6)2+(y+3)2=64(x-6)^2+(y+3)^2=64
Place xx-terms on right side.
(y+3)2=64−(x−6)2(y+3)^2=64-(x-6)^2
Take the square root of both sides.
y+3=±64−(x−6)2y+3=\pm\sqrt{64-(x-6)^2}
Subtract 3 from both sides.
y=±64−(x−6)2−3y=\pm\sqrt{64-(x-6)^2}-3
Graph both semicircles.
Y1=64−(x−6)2−3Y2=−64−(x−6)2−3\begin{aligned}Y_1&=\sqrt{64-(x-6)^2}-3 \\Y_2&=-\sqrt{64-(x-6)^2}-3 \end{aligned}
<Conic Sections>Ellipses