2.2 Circles

2.2 Circles
Chapter 2 Graphs and Functions

Concepts and Objectives
⚫ Circles
⚫ Identify the equation of a circle.
⚫ Write the equation of a circle, given the center and
the radius.
⚫ Use the completing the square method to determine
the center and radius of a circle.
⚫ Write the equation of a circle, given the center and a
point on the circle.

Circles
⚫ The geometric definition of a circle is “the set of all
points in a plane that lie a given distance from a given
point.”
⚫ We can use the distance formula to find the distance
between the center and a point:
h
k
(h, k)
(x, y)( ) ( )− + − =
2 2
2 1 2 1x x y y r
r
( ) ( )− + − =
2 2
x h y k r
( ) ( )− + − =
2 2 2
x h y k r

Circles
⚫ Example: Write the equation of a circle with its center at
(1, –2) and radius 3.

Circles
⚫ Example: Write the equation of a circle with its center at
(1, –2) and radius 3.
Let h = 1, k = –2, and r = 3. Therefore, the equation for
the circle is
( ) ( )( )− + − =−
22 2
2 31x y
( ) ( )− + + =
2 2
1 2 9x y

Circles
⚫ Example: Identify the center and radius of the circle
whose equation is
( ) ( )− + + =
2 2
7 2 81x y

Circles
⚫ Example: Identify the center and radius of the circle
whose equation is
Comparing the equation with the original form, we can
see that h is 7, k is –2, and r2 is 81 (which means that
r is 9).
Center: (7, –2) Radius: 9
( ) ( )− + + =
2 2
7 2 81x y

Sidebar: Binomial Squares
⚫ Recall that (a  b)2 expands out to
⚫ Example: Expand (2x – 5)2.
 +2 2
2a ab b
( ) ( )( )− +
2 2
2 2 2 5 5x x
2
4 20 25x x− +

⚫ This also means that anything that looks like
can be factored to (a  b)2.
⚫ Example: Factor
 +2 2
2a ab b
2
16 24 9x x+ +

⚫ This also means that anything that looks like
can be factored to (a  b)2.
⚫ Example: Factor
 +2 2
2a ab b
2
16 24 9x x+ +
2
16 4x x= 9 3= ( )( )2 4 3 24x x=
( )
2
4 3x +
✓

General Form of a Circle
⚫ The general form for the equation of a circle is
⚫ To get from the general equation back to the center-
radius form (so we can know the center and the radius),
we need to create binomial squares of both x and y.
⚫ To do this, we “complete the square” by adding numbers
to both sides of the equation that will let us make
binomial squares.
2 2
0x y cx dy e+ + + + =

⚫ Example: What is the center and radius of the circle
whose equation is
+ + − − =2 2
4 8 44 0x y x y

whose equation is
+ + − − =2 2
4 8 44 0x y x y
( ) ( )2 2
4 8 44yx yx+ + − =

whose equation is
+ + − − =2 2
4 8 44 0x y x y
( ) ( )2 2
4 8 44yx yx+ + − =
2
2
22
2
2
4 4
8
4
4
4
2
2
2
8x x y y
 
+ + 
 
   
+ + − =     
 
+ + 
 
   

whose equation is
+ + − − =2 2
4 8 44 0x y x y
( ) ( )2 2
4 8 44yx yx+ + − =
2
2
22
2
2
4 4
8
4
4
4
2
2
2
8x x y y
 
+ + 
 
   
+ + − =     
 
+ + 
 
   
( ) ( )2 2 2 2
4 8 642 4x x y y+ + − =+ +

whose equation is
The center is at (–2, 4), and the radius is 8.
+ + − − =2 2
4 8 44 0x y x y
( ) ( )2 2
4 8 44yx yx+ + − =
2
2
22
2
2
4 4
8
4
4
4
2
2
2
8x x y y
 
+ + 
 
   
+ + − =     
 
+ + 
 
   
( ) ( )2 2 2 2
4 8 642 4x x y y+ + − =+ +
( ) ( )
2 2 2
82 4x y+ =+ −

whose equation is
+ − + − =2 2
2 2 2 6 45 0x y x y

whose equation is
+ − + − =2 2
2 2 2 6 45 0x y x y
( ) ( )− + + =2 2
2 2 3 45x x y y
In order to complete the
square, the coefficients of
the square term must be 1.

whose equation is
+ − + − =2 2
2 2 2 6 45 0x y x y
( ) ( )− + + =2 2
2 2 3 45x x y y
2
2 2
22 2
1 1
2 23 45
2
3 3
2 2
222
x x y y
    
− + +
   
+ +   
   
=
 
+ +      
  

 

 
Don’t forget
to distribute!

whose equation is
+ − + − =2 2
2 2 2 6 45 0x y x y
( ) ( )− + + =2 2
2 2 3 45x x y y
2
2 2
22 2
1 1
2 23 45
2
3 3
2 2
222
x x y y
    
− + +
   
+ +   
   
=
 
+ +      
  

 

 
Don’t forget
to distribute!
   
− + + =   
   
2 2
1 3
2 2 50
2 2
x y

⚫ Example (cont.):
   
− + + =   
   
2 2
1 3
2 2 50
2 2
x y

   
− + + =   
   
2 2
1 3
2 2 50
2 2
x y
   
− + + =   
   
2 2
1 3
25
2 2
x y
Divide through
by 2.

The center is at , and the radius is 5.
   
− + + =   
   
2 2
1 3
2 2 50
2 2
x y
   
− + + =   
   
2 2
1 3
25
2 2
x y
Divide through
by 2.
 
− 
 
1 3
,
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 interesting information.
⚫ If r2 is 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 negative, then no real points will satisfy the
equation, and a graph does not exist.

Characteristics
⚫ Example: The graph of the equation
is either a circle, a point or is nonexistent. Which is it?
+ − + + =2 2
8 2 24 0x y x y

Characteristics
+ − + + =2 2
8 2 24 0x y x y
      
− + + + + = − + +               
2 2
2 28 2
8 2 24 16 1
2 2
x x y y

Characteristics
+ − + + =2 2
8 2 24 0x y x y
      
− + + + + = − + +               
2 2
2 28 2
8 2 24 16 1
2 2
x x y y
( ) ( )− + + + + = −2 2 2 2
8 4 2 1 7x x y y

Characteristics
Since r2 is negative, the graph is nonexistent.
+ − + + =2 2
8 2 24 0x y x y
      
− + + + + = − + +               
2 2
2 28 2
8 2 24 16 1
2 2
x x y y
( ) ( )− + + + + = −2 2 2 2
8 4 2 1 7x x y y
( ) ( )− + + = −
2 2
4 1 7x y

Writing the Equation of a Circle
We can now tell that the equation
is a circle with a center at (2, 3) and a radius of 4.
⚫ Suppose I wanted to know whether the point (6, 3) was
on the circle. How could I find out?
( ) ( )− + − =
2 2
2 3 16x y

We can now tell that the equation
is a circle with a center at (2, 3) and a radius of 4.
⚫ Suppose I wanted to know whether the point (6, 3) was
on the circle. How could I find out?
⚫ In order to be on the circle, the point must satisfy the
equation. That is, if we plug in 6 for x and 3 for y, and
we get 16, the point is on the circle.
( ) ( )− + − =
2 2
2 3 16x y
( ) ( )
2 2
32 166 3 ?− + − =
=16 16

⚫ We can use this idea to write the equation of a circle
given the center and a point on the circle.
⚫ Example: Write the equation of the circle with center at
(4, –5) that contains the point (–2, 3).

(–4, 5) that contains the point (–2, 3).

( ) ( )
2 2 2
4 5x y r+ − =+

( ) ( )
2 2 2
4 5x y r+ − =+
( ) ( )
2 2 2
3 542 r+− −+ =
= 2
8 r

Therefore, the equation of the circle is
( ) ( )
2 2 2
4 5x y r+ − =+
( ) ( )
2 2 2
3 542 r+− −+ =
= 2
8 r
( ) ( )+ + − =
2 2
4 5 8x y
Don’t square the
8—it’s already
squared!

Classwork
⚫ 2.2 Assignment (College Algebra)
⚫ Page 198: 6-16 (even); page 191: 34-40 (even);
page 164: 44-66 (even)
⚫ 2.2 Classwork Check
⚫ Quiz 2.1

2.2 Circles

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