How do you find the center and radius of the circle #(x-2)^2+(y+1)^2=2#?

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Answer 1 · 2016-05-29T21:55:15

The center is #(2, -1)# and the radius is #sqrt(2)#.

The circle are all the points with a fixed distance from the center. The distance is the radius.

If a point has coordinates #(x,y)# and the center #(c_x, c_y)# the distance between the point and the center is given by the Pitagora's theorem

#sqrt((x-c_x)^2+(y-c_y)^2)#

and this distance has to be equal to the radius

#sqrt((x-c_x)^2+(y-c_y)^2)=r#

In your case you do not have the square root, so you are doing the square in both sides

#(x-c_x)^2+(y-c_y)^2=r^2#.

It is enough now to compare this equation with your equation to see that

#c_x=2#, #c_y=-1# and #r^2=2#.

So the center has coordinates #(2, -1)# and the radius is #sqrt(2)#.