# One more time with Circles

• Dec 4th 2012, 07:16 PM
M670
One more time with Circles
Find the coordinates of the center and radius of the circle.
$x^2+y^2-8x+10y=-5$
So I like to rewrite it
$x^2-8x+y^2+10y=-5$
then I write it as the standard circle equation
$(x-h)^2+(y-K)^2=r^2$

So If I did this right I should have $(x-4)^2+(y+5)^2=-46$
and then my $(h,k) is (4,-5) and r=\sqrt46$
Please I need someone to confirm to me I did this properly?
Side note $x^2-8x$ is expanded to $(x-4)^2+16$ and I move the 16 over to the other side of the = sign in the equations.
• Dec 4th 2012, 07:27 PM
M670
Re: One more time with Circles
I messed this one up.... my math is wrong
• Dec 4th 2012, 07:54 PM
MarkFL
Re: One more time with Circles
You are correct to rewrite as:

$x^2-8x+y^2+10y=-5$

Now, we want to complete the square on x and y. Whatever we add to the left side, we must add to the right.

$(x^2-8x+16)+(y^2+10y+25)=-5+16+25$

Now, write everything as squares:

$(x-4)^2+(y+5)^2=6^2$

From this, we know the center is (4,-5) and the radius is 6.