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]}

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