History - Spenser, Wednesday, October 10, 2007 at 1:33pm
Find the center and radius of x^2 + y^2 - x + 2y - 3 = 0
You are looking for the form
(x-h)^2 + (y-k)^2 = r^2
x^2 -x +y^2 + 2y - 3=0
so grouping by perfect squares...
(x^2 -x + 1/4) + (y^2+2y+1)= +3-1/4-1
you do the rest. If you need help, repost.
i don't understand....
To find the center and radius of the given equation, which represents a circle, you can use the standard form of the equation of a circle, given by (x - h)^2 + (y - k)^2 = r^2. Here, (h, k) represents the center of the circle, and r represents the radius.
Let's rearrange the given equation to match the standard form:
x^2 + y^2 - x + 2y - 3 = 0
Rearranging the equation by completing the square for both x and y, you get:
(x^2 - x) + (y^2 + 2y) - 3 = 0
To complete the square for x, add (1/2)^2 = 1/4 both sides:
(x^2 - x + 1/4) + (y^2 + 2y) - 3 = 1/4
Similarly, to complete the square for y, add (1)^2 = 1 both sides:
(x^2 - x + 1/4) + (y^2 + 2y + 1) - 3 - 1 = 1/4 + 1
Simplifying the equation:
(x - 1/2)^2 + (y + 1)^2 - 3 - 1 = 5/4
(x - 1/2)^2 + (y + 1)^2 = 5/4 + 4/4
(x - 1/2)^2 + (y + 1)^2 = 9/4
Comparing the rearranged equation with the standard form, we can see that the center of the circle is (1/2, -1), and the radius is the square root of 9/4, which is 3/2.
Therefore, the center of the circle is (1/2, -1), and the radius is 3/2.