Complete the square to write the equation x^2 + y^2 - 6x - 2y + 4 = 0 in standard form. Use the standard form of the equation to identify the center and the radius of the circle.
x^2 + y^2 - 6x - 2y + 4 = 0
x^2 - 6x + 9 + y^2 - 2y + 1 = -4 + 9 + 1
(x-3)^2 + (y-1)^2 = 6
I will leave it up to you to take it from there.
To complete the square for the given equation, x^2 + y^2 - 6x - 2y + 4 = 0, we need to rearrange the terms.
1. Group the x-terms and y-terms together:
x^2 - 6x + y^2 - 2y + 4 = 0
2. Move the constant term (4) to the right side of the equation:
x^2 - 6x + y^2 - 2y = -4
3. To complete the square for the x-terms, we take half of the coefficient of x (-6) and square it: (-6/2)^2 = 9.
Add 9 to both sides of the equation:
x^2 - 6x + 9 + y^2 - 2y = -4 + 9
4. Similarly, for the y-terms, we take half of the coefficient of y (-2) and square it: (-2/2)^2 = 1.
Add 1 to both sides of the equation:
x^2 - 6x + 9 + y^2 - 2y + 1 = -4 + 9 + 1
5. Simplify the equation:
(x^2 - 6x + 9) + (y^2 - 2y + 1) = 6
6. Factor the squared terms as perfect square trinomials:
(x - 3)^2 + (y - 1)^2 = 6
Now, we have the equation in standard form, (x - h)^2 + (y - k)^2 = r^2, where (h, k) represents the center of the circle and r is the radius.
From the equation, we can determine that the center of the circle is at the point (3, 1), since h = 3 and k = 1. The radius of the circle is the square root of the constant term on the right side of the equation, which is sqrt(6).