Circles

As an application of completing the square, I'll look at equations of circles. The standard form for the equation of a circle with radius r and center $(a,b)$ is

$$(x - a)^2 + (y - b)^2 = r^2.$$

To see this, look at the right triangle below.

$$\hbox{\epsfysize=2in \epsffile{circles1.eps}}$$

The horizontal side is the difference between the x-coordinates of $(x, y)$ and $(a, b)$ , which is $|x
   - a|$ . I need to take the absolute value in case $(x, y)$ is to the left of $(a, b)$ .

Likewise, the vertical side is the difference between the y-coordinates of $(x, y)$ and $(a, b)$ , which is $|y - b|$ . I need to take the absolute value in case $(x, y)$ is below $(a, b)$ .

By Pythagoras' Theorem,

$$|x - a|^2 + |y - b|^2 = r^2.$$

Since squares are nonnegative, I can drop the absolute values and get

$$(x - a)^2 + (y - b)^2 = r^2.$$


Example. (a) Find the center and radius of the circle whose equation is

$$(x - 5)^2 + (y + 3)^2 = 36.$$

The center is $(5,-3)$ and the radius is 6.

(b) What is the equation of the circle whose center is $(-4,8)$ and whose radius is 1.1?

$$(x - (-4)^2 + (y - 8)^2 = 1.1^2, \quad\hbox{or}\quad (x + 4)^2 + (y - 8)^2 = 1.21.$$

(Don't multiply out the left side --- leave it as is.)


You can find the center and radius of a circle whose equation is not in standard form by completing the square.


Example. Find the center and radius of the circle

$$x^2 - 8 x + y^2 - 6 y = 39,$$

$$\eqalign{ x^2 - 8 x + y^2 - 6 y & = 39 \cr x^2 - 8 x + 16 + y^2 - 6 y + 9 & = 39 + 16 + 9 \cr (x - 4)^2 + (y - 3)^2 & = 64 \cr}$$

The center is $(4, 3)$ and the radius is 8.


Example. Find the center and radius of the circle

$$x^2 - 2x + y^2 + 6 y = 15,$$

To complete the square in x, I have $\dfrac{-2}{2} = -1$ , and $(-1)^2 = 1$ . To complete the square in y, I have $\dfrac{6}{2} = 3$ , and $3^2 = 9$ . So I have

$$x^2 - 2x + 1 + y^2 + 6 y + 9 = 15 + 1 + 9, \quad\hbox{or}\quad (x - 1)^2 + (y + 3)^2 = 25.$$

The center is $(1,-3)$ and the radius is $\sqrt{25} = 5$ .


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