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

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8
Jul 6, 2016

Answer:

Radius = 5
Centre (3, -4)

Explanation:

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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