# Find the center and radius of a circle

okay I know this is an extremely basic question, and I'm not sure if I'm having a brain fart. But I have this equation:

$$x^2+y^2+4y=-3$$ and I need to find the center and radius.

So I complete the square and get $$(x-0)^2+(y+2)^2=-3$$

The center is $$(0,-2)$$ but the radius obviously cannot be $$\sqrt{-3}$$

When I graph on desmos, it's showing the radius is 1. How do I find the radius?

• Check your algebra. How did you manage to complete the square on the left without adding anything to the right?
– lulu
Sep 5, 2021 at 18:35
• $x^2 + y^2 + 4y + 4 - 4 = -3$ Sep 5, 2021 at 21:14

$$x^2+y^2+4y=-3$$ is equivalent to $$x^2+(y+2)^2 = 1$$

Center = $$(0,-2)$$ Radius = $$1$$

(You missed adding 4 to the right-hand side)

• omg thank you. i made a really dumb mistake. Sep 5, 2021 at 18:47

Just like @lulu said in the comment:

If you work out your equation after completing the square:

$$(x-0)^2+(y+2)^2=-3$$,

You get $$x^2+y^2+4y+4 = -3$$

So you added $$+4$$ on the LHS, so you have to add $$+4$$ on the RHS as well, giving you $$1$$.

A generel equation of a circle can be given as $$x^2 + y^2 + 2gx + 2fy + c = 0$$ .

Where centre = $$C(-g,-f)$$ , and radius $$r = \sqrt{ g^2 + f^2 - c }$$

In your equation , $$g= 0 , f = 2 , c = 3$$ .

Therefore , $$centre = ( 0,-2)$$ and $$Radius = \sqrt{ 0^2 + {(-2)}^2 - 3} = 1$$unit