How do you graph #(x+2)^2 + (y+1)^2 =32#?

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Answer 1 · 2016-02-05T12:47:23

Circle with the center #(-2, -1)# and the radius #4sqrt(2)#.

What you have here is a circle equation.

The standard form of a circle equation is

#(x-x_m) ^2 + (y - y_m)^2 = r^2#

which describes a circle with the center point #(x_m, y_m)# and the radius #r#.

In your case, #x_m = -2#, #y_m = -1# and

#r = sqrt(32) = sqrt(16*2) = 4sqrt(2) ~~ 5.66#

Thus, you can graph a circle with the center point #(-2, -1)# and the radius #~~5.66#:

graph{(x+2)^2 + (y+1)^2 = 32 [-15.54, 16.49, -8.36, 7.66]}