SLT 5 Convert an equation of a
circle in general form,by
completing the square,to
standard form
MCPS
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Printed: October 20, 2014
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Chapter 1. SLT 5 Convert an equation of a circle in general form,by completing the square,to standard form
C HAPTER
1
SLT 5 Convert an equation
of a circle in general form,by
completing the square,to standard
form
What type of graph is produced by the equation x2 − 2x + y2 − 4y + 1 = 0? How can you turn this equation into
graphing form in order to graph it?
Watch This
Play this video beginning at 3:43.
MEDIA
Click image to the left for use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/61851
http://www.youtube.com/watch?v=g1xa7PvYV3I Conic Sections: The Circle
Guidance
The general form of any conic section is in the form Ax2 + Bxy + Cy2 + Dx + Ey + F = 0. The standard form of a
circle, (x − h)2 + (y − k)2 = r2 , is used to graph a circle in a coordinate plane since the center of the circle, (h, k),and
the radius, r, can be identified easily. To rewrite an equation of a circle in general form into an equation of a circle
in standard form, you must complete the square.
Example
The equation x2 − 2x + y2 − 4y + 1 = 0 is the equation for a circle. You can tell it is a circle because there is
both an x2 term and a y2 term, and the coefficients of each are the same. To turn this equation into the form
(x − h)2 + (y − k)2 = r2 , you must complete the square twice. Remember that when completing the square, you are
trying to figure out what number to add to make a perfect square trinomial. In order to maintain the equality of the
equation, you must add the same numbers to both sides of the equation.
x2 − 2x +
2
+ y2 − 4y +
= −1
2
x − 2x + 1 + y − 4y + 4 = −1+1 + 4
x2 − 2x + 1 + y2 − 4y + 4 = 4
(x − 1)2 + (y − 2)2 = 4
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Vocabulary
A circle is the set of all points equidistant from a given point. The general equation of a circle is (x − h)2 + (y − k)2 =
r2 .
Guided Practice
1. What is the equation of a circle centered at (−4, 5) with a radius of
√
2?
2. What are the center and radius of the circle described by the equation:
x2 − 4x + y2 + 10y + 13 = 0?
3. Graph the circle from #2.
Answers:
1. h = −4, k = 5, r =
√ 2
√
2. The equation of the circle is (x − (−4))2 + (y − 5)2 =
2 . This can be simplified to
(x + 4)2 + (y − 5)2 = 2. Notice that the signs of the 4 and the 5 in the equation are opposite from the signs of the 4
and the 5 in the center point.
2. Complete the square to rewrite the equation in graphing form:
x2 − 4x + 4 + y2 + 10y + 25 = −13+4 + 25
(x − 2)2 + (y + 5)2 = 16
The center of the circle is (2, −5) and the radius is 4.
3.
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Chapter 1. SLT 5 Convert an equation of a circle in general form,by completing the square,to standard form
Practice
Graph the following circles:
1. (x + 2)2 + (y − 5)2 = 1
2. (x − 1)2 + (y + 1)2 = 3
3. (y + 4)2 + (x − 3)2 =
1
4
4. x2 + (y − 6)2 = 8
5. (x − 2)2 + y2 = 25
Find the center and radius of the circle described by each equation.
6. x2 + 4x + y2 + 4y + 7 = 0
7. x2 − 2x + y2 + 12y + 32 = 0
8. x2 − 2x + y2 − 2y = 0
9. x2 + 8x + y2 + 9 = 0
10. x2 + 2x + y2 − 6y + 1 = 0
11. Explain why a circle with radius r that is centered at the origin will have the equation x2 + y2 = r2 .
12. Use the Pythagorean Theorem to help explain why a circle with a center in the second quadrant with radius r
and center (h, k) will have the equation (x − h)2 + (y − k)2 = r2 . Note that in the second quadrant the value for h
will be negative and the value for k will be positive.
13. Use the Pythagorean Theorem to help explain why a circle with a center in the third quadrant with radius r
and center (h, k) will have the equation (x − h)2 + (y − k)2 = r2 . Note that in the third quadrant the value for h will
be negative and the value for k will be negative.
14. Write the equation of the following circle:
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15. Write the equation of the following circle:
References
1. . . CC BY-NC-SA
2. . . CC BY-NC-SA
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