Hello I've been looking at this question for awhile but i still don't understand it. To find the centre and radius of the circle given a equation such as
(x-2)^2+(y-4)^2=8

What I've tried is
x^2-4+y^2-16=8
x^2+y^2=8+16+4
x^2+y^2 = 28

So if i wanted the center ill make either x or y =0?
which gives me the coordinates to the center being; 5.3,5.3? I'm not sure if this is correct just following what I think i know.

When you have an equation of a circle in the form

(x-a)^2+(y-b)^2=r^2

the center is at (a,b) and the radius is r. So for your equation you have a=2 and b=4, so the center is at (2,4), and radius = sqrt(8).

Also,when you expand (x-2)^2 you should get (x-2)(x-2) = x^2-4x+4.

Quote:

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Originally Posted by ebaines

When you have an equation of a circle in the form

(x-a)^2+(y-b)^2=r^2

the center is at (a,b) and the radius is r. So for your equation you have a=2 and b=4, so the center is at (2,4), and radius = sqrt(8).

Also,when you expand (x-2)^2 you should get (x-2)(x-2) = x^2-4x+4.

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just to confirm the coordinates for the center of the following my equation is negative, so does that mean it should be (-2,-4)?

Quote:

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Originally Posted by aztect

just to confirm the coordinates for the center of the following my equation is negative, so does that mean it should be (-2,-4)?

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No. A circle with center at (-2,4) and radius would have the equation .