Standard Form of Equation of Circle

[nycmath]

Precalculus
Michael Sullivan
Edition 10
Chapter 1, Section 1.4

Find the standard form of the equation of each circle.

36. Center (1, 0) and containing the point (-3, 2).

40. With endpoints of a diameter at (4, 3) and (0, 1).

[Math Tutor]

For both of these you just need the center and the radius, then plug into [tex](x - h)^2 + (y - k)^2 = r^2[/tex].

36. The center is [tex](1, 0)[/tex] and the circle passes through [tex](-3, 2)[/tex], so the radius is the distance between those two points:
[tex]r^2 = (-3 - 1)^2 + (2 - 0)^2 = 16 + 4 = 20[/tex]
No need to take the square root, since the formula wants [tex]r^2[/tex] anyway.
Answer: [tex](x - 1)^2 + y^2 = 20[/tex]

40. The endpoints of a diameter are given, so the center is just the midpoint:
[tex]\left(\frac{4 + 0}{2}, \frac{3 + 1}{2}\right) = (2, 2)[/tex]
The radius is the distance from the center to either endpoint:
[tex]r^2 = (4 - 2)^2 + (3 - 2)^2 = 4 + 1 = 5[/tex]
Answer: [tex](x - 2)^2 + (y - 2)^2 = 5[/tex]

Quick sanity check on #40: the full diameter has length [tex]\sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}[/tex], so [tex]r = \sqrt{5}[/tex], which matches.