How do you graph #(x+2)^2 + (y+1)^2 =32#?
1 Answer
Feb 5, 2016
Circle with the center
Explanation:
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
In your case,
#r = sqrt(32) = sqrt(16*2) = 4sqrt(2) ~~ 5.66#
Thus, you can graph a circle with the center point
graph{(x+2)^2 + (y+1)^2 = 32 [-15.54, 16.49, -8.36, 7.66]}