How do you find the center and radius of the circle ##x^2 + y^2 - 6x + 8y = 0##?

Radius = 5 Centre (3, -4)

The general equation of a circle is ##x^2+y^2=r^2##

Rearranging the equation ##x^2-6x+y^2+8y=0##

Then, you would want to find the perfect squares of ##x## and ##y## remembering to balance the equation ##x^2-6xcolor(red)(+9)+y^2+8ycolor(blue)(+16)=0##

And because you added something that wasn't there before, you have to take it away again ##x^2-6xcolor(red)(+9-9)+y^2+8ycolor(blue)(+16-16)=0##

Then simplifying it to: ##(x-3)^2color(red)(-9)+(y+4)^2color(blue)(-16)=0##

##(x-3)^2## comes from ##x^2-6x+9## ##(y+4)^2## comes from ##y^2+8y+16##

If we rearrange the equation like this: ##(x-3)^2+(y+4)^2color(red)(-9)color(blue)(-16)=0## And simplify it to: ##(x-3)^2+(y+4)^2-25=0##

We can determine the centre and radius by:

##(x-3)^2+(y+4)^2-25=0## Adding 25 to both sides ##(x-3)^2+(y+4)^2=25##

And comparing it to ##(x-h)^2+(y-k)^2=r^2## The radius of the equation above would be at ##(h, k)## with a radius of ##r##

Therefore, your centre would have the coordinates ##(3, -4)## and have a radius of ##sqrt25## which equals to 5