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QUESTION

# 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