Distance, Circles, And Quadratic Equations Worksheets Page 4
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9H4
Appendix H: Distance, Circles, and Quadratic Equations
Still another version of the equation of a circle can be obtained by multiplying both sides
of (4) by a nonzero constant A. This yields an equation of the form
2
+ Ay
2
+ Dx + Ey + F = 0
(5)
Ax
where A, D, E, and F are constants and A = 0.
If the equation of a circle is given by (4) or (5), then the center and radius can be found by
first rewriting the equation in standard form, then reading off the center and radius from that
equation. The following example shows how to do this using the technique of completing
the square. In preparation for the example, recall that completing the square is a method
for rewriting an expression of the form
+ bx
2
x
as a difference of two squares. The procedure is to take half the coefficient of x, square it,
and then add and subtract that result from the original expression to obtain
+ bx = x
+ bx + (b / 2)
− (b / 2)
= [x + (b / 2)]
− (b / 2)
2
2
2
2
2
2
x
Example 7
Find the center and radius of the circle with equation
+ y
− 8x + 2y + 8 = 0
+ 2y
+ 24x − 81 = 0
2
2
2
2
(a) x
(b) 2x
Solution (a).
First, group the x-terms, group the y-terms, and take the constant to the
right side:
2
− 8x) + (y
2
+ 2y) = −8
(x
Next we want to add the appropriate constant within each set of parentheses to complete the
square, and subtract the same constant outside the parentheses to maintain equality. The
appropriate constant is obtained by taking half the coefficient of the first-degree term and
squaring it. This yields
− 8x + 16) − 16 + (y
+ 2y + 1) − 1 = −8
2
2
(x
from which we obtain
(x − 4)
+ (y + 1)
= −8 + 16 + 1 or (x − 4)
+ (y + 1)
= 9
2
2
2
2
Thus from (3) the circle has center (4, −1) and radius 3.
The given equation is of form (5) with A = 2. We will first divide through
Solution (b).
by 2 (the coefficient of the squared terms) to reduce the equation to form (4). Then we will
proceed as in part (a) of this example. The computations are as follows:
+ y
+ 12x −
= 0
81
2
2
x
We divided through by 2.
2
+ 12x) + y
=
81
2
2
(x
2
+ 12x + 36) + y
=
+ 36
81
2
2
(x
We completed the square.
2
(x + 6)
+ y
=
153
2
2
2
153
From (3) the circle has center (−6, 0) and radius
.
2
DEGENERATE CASES OF A CIRCLE
There is no guarantee that an equation of form (5) represents a circle. For example, suppose
that we divide both sides of (5) by A, then complete the squares to obtain
(x − x
+ (y − y
= k
2
2
)
)
0
0
Depending on the value of k, the following situations occur:
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