General Form of Equation of Circle

[nycmath]

Precalculus
Michael Sullivan
Edition 10
Chapter 1, Section 1.4

Find the radius r, the center (h, k) and the intercepts, if any. Graph the circle.

28. x^2 + y^2 - 6x + 2y + 9 = 0

Hint: Complete the square

[Math Tutor]

Group the x terms and the y terms first, then complete the square on each:

[tex]x^2 - 6x + y^2 + 2y = -9[/tex]

Half of [tex]-6[/tex] is [tex]-3[/tex], and [tex](-3)^2 = 9[/tex]. Half of [tex]2[/tex] is [tex]1[/tex], and [tex]1^2 = 1[/tex]. Add both to each side:

[tex]x^2 - 6x + 9 + y^2 + 2y + 1 = -9 + 9 + 1[/tex]
[tex](x - 3)^2 + (y + 1)^2 = 1[/tex]

So the center is [tex](h, k) = (3, -1)[/tex] and the radius is [tex]r = 1[/tex].

Intercepts:

x-intercepts, set [tex]y = 0[/tex]:
[tex](x - 3)^2 + 1 = 1 \Rightarrow (x - 3)^2 = 0 \Rightarrow x = 3[/tex]
Only one, at [tex](3, 0)[/tex]. That is a double root, which makes sense because the circle is tangent to the x-axis there.

y-intercepts, set [tex]x = 0[/tex]:
[tex]9 + (y + 1)^2 = 1 \Rightarrow (y + 1)^2 = -8[/tex]
No real solution, so there are no y-intercepts. Also obvious from the picture, since the center is 3 units from the y-axis but the radius is only 1.

For the graph: plot the center at [tex](3, -1)[/tex] and go 1 unit in each direction, so the circle passes through [tex](2, -1)[/tex], [tex](4, -1)[/tex], [tex](3, 0)[/tex] and [tex](3, -2)[/tex]. It sits entirely in the fourth quadrant except for that single touch point on the x-axis.