Center and radius of x^2 + y^2 – 4y = 0.
STD Form of Circle:
(x - h)^2 + (y - k)^2 = r^2, C(h , k).
x^2 + y^2 - 4y = o,
Complete the square:
x^2 + y^2 - 4y + (-4 / 2)^2 = (-4/2)^2,
x^2 + y^2 - 4y + 4 = 4,
Convert the perfect sq to a binomial:
(x - 0 )^2 + (y - 2)^2 = 4 = r^2, r=2.
C(0 , 2).
To find the center and radius of the given equation, which represents a circle, we need to write the equation in the standard form: (x - h)^2 + (y - k)^2 = r^2, where (h, k) represents the center of the circle and r represents the radius.
Let's start with the given equation: x^2 + y^2 – 4y = 0.
First, we complete the square to isolate the variables x and y.
Rearranging the terms, we get: x^2 + (y^2 - 4y) = 0.
To complete the square for the y terms, we take half the coefficient of y (-4) and square it, which gives us (-2)^2 = 4. We add this value to both sides of the equation:
x^2 + (y^2 - 4y + 4) = 4.
Now, the y terms can be factored as a perfect square: (y - 2)^2.
Our equation becomes: x^2 + (y - 2)^2 = 4.
Comparing this equation to the standard form, we see that the center is located at (0, 2), as (h, k) = (0, 2). The radius of the circle is given by r = √4 = 2.
Therefore, the center of the circle is (0, 2), and its radius is 2 units.