Question

find the center and radius of a circle given the equation in general form

x^2+y^2+4x-6y+2=0.

center: ?
radius: ?

Answer 1

Please use the same name for all of your posts.

By using different names, you're usurping names that are already used by other students.

Answer 2

I am not sure what you're talking about.

Answer 3

Meghan, John, and Jen all have the same ip address, a computer in St. Louis, Missouri.

Answer 4

Rewrite the equation in the form

(x + 2)^2 + (y - 3)^3 = 11

This shows that it is a circle centered at x = -2, y = 3.
See of you can guess the radius, from the constant term on the right.

Answer 5

[AI-generated answer — Explain Bot]

To find the center and radius of a circle given its equation in general form, we need to rewrite the equation in standard form. The standard form of the equation of a circle is (x - h)^2 + (y - k)^2 = r^2, where (h, k) represents the coordinates of the center, and r represents the radius.

Let's start by rearranging the given equation:
x^2 + y^2 + 4x - 6y + 2 = 0

First, we group the x-terms and y-terms:
(x^2 + 4x) + (y^2 - 6y) + 2 = 0

Next, we complete the square for both the x and y terms separately. To complete the square for the x-terms, we add the square of half the coefficient of x, which is (4/2)^2 = 4:
(x^2 + 4x + 4) + (y^2 - 6y) - 4 + 2 = 0

Similarly, for the y-terms, we add the square of half the coefficient of y, which is (-6/2)^2 = 9:
(x^2 + 4x + 4) + (y^2 - 6y + 9) - 4 + 2 = 9

Now, we simplify the equation:
(x + 2)^2 + (y - 3)^2 = 9

Comparing this equation with the standard form, we can see that the center is (-2, 3) and the radius is the square root of 9, which is 3.

Therefore, the center of the circle is (-2, 3) and the radius is 3.