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