Equations of Circles
(x – a)2 + (y – b)2 = r2
(a, b) = center of the circle r = radius of the circle
(x – 4)2 + (y + 3)2 = 36
(4, -3) = center of the circle
6 = radius of the circle
What is the center and the radius of this circle?
Let’s Look at the Graph of
(x – 4)2 + (y + 3)2 = 36
center of the circleis (4, -3)
radiusof the circleis 6
Equations of an Ellipse
The x-intercepts a
2 2a cbx y Standard Form
The y-intercepts bThe center is the origin
2 23616 576x y
The x-intercepts 6
The y-intercepts 4
The center is the origin
Intercept Form22
22 1yxba
The x-intercepts aThe y-intercepts b
The center is the origin22
22 461
yx
22
1361
6yx
The x-intercepts 6
The y-intercepts 4
The center is the origin
Let’s Look at the Graph of
16x2 + 36y2 = 576
center ofthe ellipse
is the origin
x-interceptsare 6
y-interceptsare 4
Rewriting the Equation of an Ellipse
Rewrite the following equation of an ellipse in intercept form.
2 21625 400x y Since the equation must be equal to 1, divide by 400.
222 400400 4
16500 400yx
Simplify the fractions.
22
2161
5yx I think I can do this
without having to do too much work.
Rewrite the following equation of an ellipse in standard form. 22
811
4yx
Multiply by the common denominator.
22
4324 324 2
811 3 4
yx
Simplify the equation.
2 28 3 414 2x y
Just switch the coefficients of x and
y.
Graphing Circles and Ellipses 2 2( ) ( 15)3 6yx
Graph the following circle.Write the equation of the circle whose graph looks like this.
-8 -6 -4 -2 2 4 6 8
8
6
4
2
-2
-4
-6
-8
2 234) 25( ( )yx
Graph the following ellipse.
2 22 2 559 2x y
-8 -6 -4 -2 2 4 6 8
8
6
4
2
-2
-4
-6
-8
22
141
8yx
Write the equations of the following two ellipses in both standard form and intercept form.
-8 -6 -4 -2 2 4 6 8
8
6
4
2
-2
-4
-6
-8
2 24 12 5925 2x y 22
2491
5yx
2 21610 16000x y 22
0161
10yx
Equations of a Hyperbola
xy = kAs x increases, y decreases so that the product of x and y is
always k
xy = 8As x increases, y decreases so that the product of x and y is
always 8
•A hyperbola is a function•The coordinate axes are asymptotes of the graph•When k > 0, the graph is in quadrant I and quadrant III•When k < 0, the graph is in quadrant II and quadrant IV•Each branch of the hyperbola is the reflection of the other in the origin
Let’s Look at the Graph of
xy = 8Since k > 0,
the graph is in quadrant I
and quadrant III
The x and y axes are
asymptotes of the graph
Let’s Look at the Graph of
xy = -8Since k < 0,
the graph is in quadrant II
and quadrant IV
The x and y axes are
asymptotes of the graph
Graphing Hyperbolas
6xy Solve the equation for y. 6y
x
Make a table of values.
1
2
3
6
12
24
x y6
3
2
1
.5
.25
Since k is positive,
the graph is
in quadrants I and III
4xy
4y
x
x y1
2
4
8
16
32
4
2
1
.5
.25
.125
Since k is negative,
the graph is
in quadrants II and
IV
More Graphing Hyperbolas
12xy Solve the equation for y. 12y
x
Make a table of values.
1
2
3
6
12
24
x y12
6
4
2
1
.5
Since k is positive,
the graph is
in quadrants I and III
24xy
24y
x
x y1
2
3
4
6
8
12
24
24
12
8
6
4
3
2
1
Since k is positive,
the graph is
in quadrants I and III
6xy 6xy
12xy
6xy
12xy
24xy
6xy
12xy
24xy
72xy