Conic Sections – The Circle
Objective: To define a circle and work with the equation of a circle in general and
standard form.
Development:
1. Definitions
a. A circle is the set of all points in a plane which are equidistant from a
fixed point. The fixed point is called the center. The equal length that all
points are from the center is called the radius.
Draw a diagram of a circle with a radius of 3 cm.
2. The distance formula can be used to determine the equation of a circle.
If the center was at the origin:
d= ( x  x1 ) 2  ( y2  y1 ) 2
r=
( x  0) 2  ( y  0) 2
r2 = x2 + y2
If the center is moved from the origin to (h,k) then,
r=
( x  h) 2  ( y  k ) 2
r2 = (x – h)2 + (y – k)2
Left in this form, the equation is called STANDARD FORM.
Eg. Write the equation of a circle with a radius of 8 units that has been translated 5
units to the right and 1 unit down from its standard position (ie. Center at origin).
r2 = (x – h)2 + (y – k)2
82 = (x – 5)2 + (y + 1)2
64 = (x – 5)2 + (y + 1)2
This is called standard form.
GENERAL FORM results when standard form is expanded.
Eg. If (x – 5)2 + (y + 1)2 = 64
x2 – 10x + 25 + y 2 + 2y + 1 = 64
x2 – 10x + y2 + 2y + 26 = 64
x2 + y2 – 10x + 2y – 38 = 0
This equation is in general form.
Remember, general form is written as
Ax2 + Cy2 + Dx + Ey + F = 0
Examples.
1. Write the equation of a circle with a center at (-6,-9) and a radius of 5 3 .
(x + 6)2 + (y + 9)2 = 75
x - intercepts: none
y - intercepts: y = - 9 
39
{ d i d i}
{y d9  5 3i  y  d9  5 3i}
Domain: x 6  5 3  x  6  5 3
Range:
If the radius of a circle is not given, it will need to be determined before an equation
can be written.
2. Write the equation of a circle with a center at (-3,6) which passes through
(5,-1).
r is the distance from C(-3,6) to P(5,1) [ 113 ]
Equation is: (x + 3)2 + (y – 6)2 = 113
x-intercepts: - 3  77
y-intercepts: 6  104
{x d3  113i  x  d3  113i}
{y d6  113i  x  d6  113i}
domain:
range:
NOTE: If the equation of a circle is written in standard form, its center and radius
can easily be read.
3. Sketch the graph of (x +8)2 + (y – 2)2 = 36
Compare to (x – h)2 + (y – k)2 = r2
h = -8, k = 2
Centre (-8,2)
Radius= 6
Domain: x  8  6   x   8  6 
Range:


 y  2  6  y   2  6
x-intercepts: - 8  7
y-intercepts: none
4. (x – 4)2 + (y + 12)2 = 62 (Sketch the graph)
Centre (4,-12)
Radius = 62
Remember, if an equation is written in general form, we can’t read the center and
radius from it.
Example: Find the center and radius of x2 + y2 + 6x –2y – 26 = 0 (include
domain and range, x-intercepts, y-intercepts)
If this had been in standard form, the center and radius would have been
easy to read. To convert from general form to standard form, we use
steps called COMPLETING THE SQUARE.
x2 + y2 + 6x – 2y – 26 = 0
a. Group “x” terms and “y” terms together and isolate the constant.
(x2 + 6x) + (y2 – 2y) = 26
b. In the “x” bracket, a third term is added to create a perfect square
trinomial. To choose this third term, take the coefficient of x, divide
by 2, and square. Then, do the same thing for y.
(x2 + 6x + 9) + (y2 – 2y +1) = 26
Balance the equation by adding the same amount to the other side.
(x2 + 6x + 9) + (y2 – 2y + 1) = 26 + 9 + 1
c. Factor and simplify
(x + 3)2 + (y – 1)2 = 36
d. Read center and radius
Center (-3,1)
Radius = 6
Domain: x 9  x  3
Range:
m
r
y 5  y  7r
m
x-intercepts: - 8  7
y-intercepts: 1  27
5. Find the center and radius of
4x2 + 4y2 + 20x – 4y –74 = 0
(4x2 + 20x) + (4y2 – 4y) = 74
At this point, factor out the coefficient of the x2 and y2 from each bracket.
4(x2 + 5x) + 4(y2 – y) = 74
4(x2 + 5x +
1
25
) + 4(y2 – y +
) = 74
4
4
When balancing, remember to multiply by the factored out coefficient.
4(x2 + 5x +
4(x +
(x +
5
1
) + 4(y2 – y +
) = 74 + 25 + 1
2
4
5 2
1 2
) + 4(y ) = 100
2
2
5 2
1 2
) + (y ) = 25
2
2
5 1
Center ( ,
)
2 2
Radius = 5
15
5
Domain: x   x 
2
2
Range:
R
U
S
V
T
W
9
11U
R
y  y V
S
T 2 2W