Distance, Circles, And Quadratic Equations Worksheets Page 3
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9Appendix H: Distance, Circles, and Quadratic Equations
H3
(x, y) will lie on this circle if and only if
(x − x
+ (y − y
= r
2
2
)
)
0
0
or, equivalently,
(x − x
+ (y − y
= r
2
2
2
(3)
)
)
0
0
This is called the standard form of the equation of a circle.
Example 4
Find an equation for the circle of radius 4 centered at (−5, 3).
= −5, y
= 3, and r = 4 we obtain
Solution.
From (3) with x
0
0
(x + 5)
2
+ (y − 3)
2
= 16
If desired, this equation can be written in an expanded form by squaring the terms and then
simplifying:
+ 10x + 25) + (y
− 6y + 9) − 16 = 0
2
2
(x
+ y
+ 10x − 6y + 18 = 0
2
2
x
Find an equation for the circle with center (1, −2) that passes through
Example 5
(4, 2).
The radius r of the circle is the distance between (4, 2) and (1, −2), so
Solution.
r =
(1 − 4)
+ (−2 − 2)
= 5
2
2
We now know the center and radius, so we can use (3) to obtain the equation
(x − 1)
2
+ (y + 2)
2
= 25 or x
2
+ y
2
− 2x + 4y − 20 = 0
FINDING THE CENTER AND RADIUS OF A CIRCLE
When you encounter an equation of form (3), you will know immediately that its graph is a
circle; its center and radius can then be found from the constants that appear in the equation:
(x − x
+
(y − y
=
2
2
2
)
)
r
0
0
x-coordinate of the center is x
.
y-coordinate of the center is y
.
radius squared
0
0
Example 6
y
+ y
= 1
2
2
x
equation of a circle
center
radius r
(x
, y
)
0
0
1
(x − 2)
2
+ (y − 5)
2
= 9
(2, 5)
3
x
(x + 7)
2
+ (y + 1)
2
= 16
(−7, −1)
4
+ y
= 25
2
2
x
(0, 0)
5
√
(x − 4)
2
+ y
2
= 5
(4, 0)
5
+ y
= 1, which is centered at the origin and has radius 1, is of special
2
2
The unit circle
The circle x
importance; it is called the unit circle (Figure H.6).
Figure H.6
OTHER FORMS FOR THE EQUATION OF A CIRCLE
An alternative version of Equation (3) can be obtained by squaring the terms and simplifying.
This yields an equation of the form
+ y
+ dx + ey + f = 0
2
2
(4)
x
where d, e, and f are constants. (See the final equations in Examples 4 and 5.)
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