6-1 Circles: Unit 6 Conics
6-1 Circles: Unit 6 Conics
6-1 Circles: Unit 6 Conics
Unit 6 Conics
Concepts and Objectives
Circles (Obj. #19)
Identify the equation of a circle.
( x2 − x1 ) + ( y2 − y1 ) = r
2 2
(x, y)
r
( x − h) + ( y − k ) = r
2 2 k
(h, k)
( ) ( )
2 2
x − h + y − k = r 2
h
Circles
Example: Write the equation of a circle with its center at
(1, –2) and radius 3.
( ) ( ) =9
2 2
x − 1 + y + 2
Graphing Circles
To graph a circle on graph paper, plot the center point
and count out the radius. Open the compass to that
point and draw the circle.
Example: Graph ( x − 3) + ( y − 4 ) = 9
2 2
Graph ( x − 3) + ( y − 4 ) = 9
2 2
( y − 4 ) = 9 − ( x − 3)
2 2
y − 4 = 9 − ( x − 3)
2
y = 9 − ( x − 3) + 4
2
Since you can only graph functions, you will also need to
enter the negative of this for the lower half of the circle.
Graphing Circles
Enter y = 9 − ( x − 3) + 4
2
Enter y = − 9 − ( x − 3) + 4
2
x 2 − 2hx + h2 + y 2 − 2ky + k 2 − r 2 = 0
( )
x 2 + y 2 + ( −2h) x + ( −2k ) y + h2 + k 2 − r 2 = 0
If we let c = –2h, d = –2k, and e = h2 + k2 – r2, we have
x 2 + y 2 + cx + dy + e = 0
General Form of a Circle
To get from the general equation back to the center-
radius form (so we can know the center and the radius),
we complete the square for both x and y.
x + 4 x + + y − 8 y + = 44 + 4 + 16
2
2 2
( ) ( )
x 2 + 4 x + 22 + y 2 − 8 y + 42 = 64
( x + 2) + ( y − 4 ) = 82
2 2
2
1
2
3
2
1 9
2 x − x + + 2 y + 3 y + = 45 + +
2
2 2 2 2
2 2
1 3
2 x − + 2 y + = 50
2 2
General Form of a Circle
Example (cont.):
2 2
1 3
2 x − + 2 y + = 50
2 2 Divide through
by 2.
2 2
1 3
x − + y + = 25
2 2
1 3
The center is at , − , and the radius is 5.
2 2
Characteristics of r2
When we convert from the general form to the center-
radius form, the constant on the right-hand side tells us
some basic information.
If r2 is > 0 (positive), the graph of the equation is a
circle with radius r.
If r2 is equal to 0, the graph of the equation is a single
point (h, k).
If r2 is < 0 (negative), then no real points will satisfy
the equation, and a graph does not exist.
Characteristics
Example: The graph of the equation
x 2 + y 2 − 8 x + 2 y + 24 = 0
is either a point or is nonexistent. Which is it?
2 8
2
2
2
x − 8 x + + y + 2 y + = −24 + 16 + 1
2
2 2
( ) ( )
x 2 − 8 x + 42 + y 2 + 2 y + 12 = −7
( x − 4 ) + ( y + 1) = −7
2 2
( x + 4 ) + ( y − 5) = r 2
2 2
( ) ( )
2 2
− 2 + 4 + 3 − 5 = r 2
8 = r2
Therefore, the equation of the circle is
( x + 4 ) + ( y − 5)
2 2
=8
Don’t square the
8—it’s already
squared!
Circular Inequalities
Circular inequalities are fairly straightforward. For the
center-radius form of the circle, the graph will be
The region inside the circle if the symbol is “<” or “≤”
< ≥
Homework
College Algebra (brown book)
Page 199: 21-30 (×3s)