Let's review what we already know about circles.
Definition:
A
circle is a
locus (set) of points in a plane equidistant from a fixed point.
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Circle
whose center is at the origin |
Circle
whose center is at (h,k)
(Called the "center-radius
form") |
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Equation:
Example:
Circle with center (0,0), radius 4
Graph:
 |
Equation:
Example: Circle with
center (2,-5), radius 3
Graph:
 |
Now, if we "multiply out" the above example
we will get:
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When multiplied out, we obtain
the
"standard form"
of the equation of
a circle. Notice that in standard form we can clearly see
that the equation of a circle has both x2
and y2 terms and these terms have the same coefficient (usually 1).
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When the equation of a circle appears in "standard form", it is often
beneficial to convert the equation to "center-radius" form to easily
read the center coordinates and the radius.
Examples:
1.
Convert
into center-radius form. |
This
conversion requires use of the technique of
completing
the square.
 |
• Start by grouping the x related terms together and the y
related terms
together. Take any numerical constants (plain numbers) to the right side.
• Set up boxes to receive your values for creating perfect
square trinomials. Balance both sides.
• Find the missing value by taking have of the "middle term" and
squaring. This value will always be positive.
• Rewrite in factored form. |
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You can now read that the
center of the circle is at (2, 3) and the radius is
 . |
2.
How do the coordinates of the center of a circle relate
to C and
D when the equation of
the circle is in the standard form
? |
Let's make some observations.
Re-examine our previous equations in standard form and center-radius
form. Do you see a relationship between the center coordinates and
C and D?
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Standard form |
Center-radius form |
C = -4,
D = 10 |
Center (2, -5) |
C = -4,
D = -6 |
Center (2, 3) |
|
 |
It
appears that the values of C and D
are (-2) times the coordinates of the center respectively.
Why is this occurring?
When  is
expanded,  becomes
 , where the center
term's coefficient doubles the value of
-2.
Remember that while the equation deals with
, the actual x-coordinate
of the center of this circle is
+2.
(Read more about these
relationships at the Resource Page.)
| 3.
Write the equation of a circle whose diameter has
endpoints (4, -1) and (-6, 7). |
 |
Find the center by
using the midpoint formula.
Find the radius by using the distance
formula.
Points (-6,7) and (-1,3)
were used here. (d =
distance, or radius)
 |
Equation:  |
| 4.
Write the equation for the circle shown below if it is
shifted 3 units to the right and 4 units up. |
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A shift of 3 units to the right and 4 units up places the center
at the point (3, 4). The radius of the circle can be seen
from the graph to be 5.
Equation:
 |
5.
Convert
 into
center-radius form. |
|
 |
Whoa!!! This
equation looks different. Are we sure this is a
circle??? In this equation, both
the x and y terms appear in
squared form and their
coefficients (the numbers in front
of them) are the same. Yes, we have a
circle here! We will, however, have to deal with
the coefficients of 2 before we can complete the square. |
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Center: (-3/2, 2)
Radius: 1/2 |
• group the terms •
divide through by 2
• get ready to create perfect squares
• take half of the "middle term" and square it
• factor and write in center-radius form
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