Final answer:
The standard form of the equation "x^ 2 + y^ 2 + 4x -8y+ 13 =0" is (x + 2)^2 + (y - 4)^2 - 8 = 0.
Step-by-step explanation:
The given equation is x^2 + y^2 + 4x - 8y + 13 = 0.
We can rewrite the equation in standard form by completing the square for both the x and y terms.
To do this, we need to add and subtract the square of half the x coefficient and the square of half the y coefficient:
x^2 + 4x + 4 - 4 - 8y + 16 - 13 = 0
(x + 2)^2 - 4 + (y - 4)^2 - 4 = 0
(x + 2)^2 + (y - 4)^2 - 8 = 0
This is now in standard form, which is (x - h)^2 + (y - k)^2 = r2.
In this case, the center of the circle is (-2, 4) and the radius is √8.