James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

appendix C Graphs of Second-Degree Equations A17
y
r
P(x, y)
from the center Csh, kd is r. (See Figure 1.) Thus P is on the circle if and only if | PC | − r.
From the distance formula, we have
ssx 2 hd 2 1 sy 2 kd 2 − r
C(h, k)
or equivalently, squaring both sides, we get
sx 2 hd 2 1 sy 2 kd 2 − r 2
0
FIGURE 1
x
This is the desired equation.
1 Equation of a Circle An equation of the circle with center sh, kd and
radius r is
sx 2 hd 2 1 sy 2 kd 2 − r 2
In particular, if the center is the origin s0, 0d, the equation is
x 2 1 y 2 − r 2
Example 1 Find an equation of the circle with radius 3 and center s2, 25d.
SOLUTIOn From Equation 1 with r − 3, h − 2, and k − 25, we obtain
sx 2 2d 2 1 sy 1 5d 2 − 9
n
Example 2 Sketch the graph of the equation x 2 1 y 2 1 2x 2 6y 1 7 − 0 by first
showing that it represents a circle and then finding its center and radius.
SOLUTIOn We first group the x-terms and y-terms as follows:
sx 2 1 2xd 1 sy 2 2 6yd − 27
Then we complete the square within each grouping, adding the appropriate constants
(the squares of half the coefficients of x and y) to both sides of the equation:
sx 2 1 2x 1 1d 1 sy 2 2 6y 1 9d − 27 1 1 1 9
or sx 1 1d 2 1 sy 2 3d 2 − 3
Comparing this equation with the standard equation of a circle (1), we see that h − 21,
k − 3, and r − s3 , so the given equation represents a circle with center s21, 3d and
radius s3 . It is sketched in Figure 2.
y
(_1, 3)
FIGURE 2
x 2 1 y 2 1 2x 2 6y 1 7 − 0
0
1
x
n
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