How do you find the center and radius of the circle given #x^2+(y+2)^2=4#?

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Answer 1 · 2016-11-24T17:06:17

Please see the explanation.

The standard form for the equation of a circle is:

#(x - h)^2 + (y - k)^2 = r^2#

where #(x, y)# is any point on the circle, #(h, k)# is the center, and r is the radius.

I suggest that you modify any given equation into this form and then report the easily observed values.

Change the + 2 to - -2:

#x^2 + (y - -2)^2 = 4#

Insert a -0 in the x term:

#(x - 0)^2 + (y - -2)^2 = 4#

Write the constant as a square:

#(x - 0)^2 + (y - -2)^2 = 2^2#

It is easily observed that the center is #(0, -2)# and the radius is 2.