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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

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