Question

what is the radius of a circle with the equation
x^2+y^2+6x-2y+3=0 (round answer to the nearest thousandth)

Answer 1

\[\sqrt{7}\]

Answer 2

would you mind showing your work? :)

Answer 3

x^2+y^2+6x-2y+3 = (x+3)^2 + (y-1)^2 -9 -1 +3
So we have equation (x+3)^2 + (y-1)^2 = 7. It implies that R^2 = 7.

Answer 4

oh I see thanks

Answer 5

\[x^2+y^2+6x-2y+3=0\]You're going to complete the square twice, once for \(x\) and once for \(y\).
\[(x^2+6x) + (y^2-2y) + 3 = 0\]I like to add and subtract the same value from the side where I'm completing the square rather than adding to both sides.

We take half of the coefficient of \(x\), and square it: \(\frac{1}{2}*6 = 3\), \(3^2 = 9\). Now we both add the 9 inside the parentheses for the \(x\) group, and subtract it outside:
\[(x^2+6x+9) + (y^2-2y) + 3 - 9 = 0\]Now we can rewrite the \(x\) group as\[(x+3)^2+(y^2-2y)-6=0\]Repeat the process with \(y\):

Take half of the coefficient of \(y\), and square it: \(\frac{1}{2}*(-2) = -1\), \((-1)^2=1\). Add inside the parentheses for the \(y\) group and subtract it outside:
\[(x+3)^2 + (y^2-2y+1) - 6 - 1 = 0\]\[(x+3)^2+(y-1)^2 - 7 = 0\]\[(x+3)^2+(y-1)^2=7\]\[(x+3)^2+(y-1)^2 = (\sqrt{7})^2\]So, we have a circle with center at \((-3,1)\) and radius \(\sqrt{7} \approx 2.646\)