Find the center and radius of a circle from its equation

Key takeaway: When a circle equation is not already in standard form, completing the square is the fastest way to reveal the center and radius. The key is to group the $y$-terms carefully.

Given

$x^2 + y^2 + 6y + 7 = 0$

Rewrite it by moving the constant:

$x^2 + y^2 + 6y = -7$

Now complete the square for the $y$-terms.

Take half of 6, which is 3, and square it to get 9.

Add 9 to both sides:

$x^2 + y^2 + 6y + 9 = 2$

Now factor the perfect square:

$x^2 + (y+3)^2 = 2$

This is the standard form of a circle:

$(x-h)^2 + (y-k)^2 = r^2$

So the center is

$\boxed{(0,-3)}$

and the radius is

$\boxed{\sqrt{2}}$

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Pitfalls the pros know 👇 A frequent mistake is forgetting that the $x^2$ term already matches a perfect square, so there is no $x$-term to complete. Another one is adding 9 only to the left side; that breaks the equation. Whatever you add to complete the square must be added to both sides.

What if the problem changes? If the equation were

$x^2+y^2+6y-5=0,$

then the same process would give

$x^2+(y+3)^2=14,$

so the center would still be $(0,-3)$, but the radius would become $\sqrt{14}$. The center comes from the linear terms, while the radius depends on the constant.

Tags: standard form, completing the square, radius