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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?

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

3 Answers 3

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

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

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

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  • $\begingroup$ omg thank you. i made a really dumb mistake. $\endgroup$ Sep 5, 2021 at 18:47
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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 $1$.

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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

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