How do you graph the circle #x^2+y^2+3x+4y+4=0#?

1 Answer
Jan 25, 2016

Answer:

# (x - 3/2)^2 + (y - 2)^2 = 9/4 #
graph{(x-3/2)^2 + (y-2)^2 - 9/4 = 0 [-2.205, 5.55, -0.087, 3.954]}

Explanation:

#x^2 + y^2 + 3x + 4y + 4 = 0 #

Step 1. Group x and y terms together.
#(x^2 + 3x + alpha) + (y^2 + 4y + beta) = -4 + alpha + beta #
Now the goal to build a perfect by setting
#alpha = (3/2)^2 = 9/4; beta = (4/2)^2 = 4#

Now the circle equation in standard form is:
# (x-h)^2 + (y-k)^2 = r^2 #
#h = sqrt(alpha) = 3/2; k = sqrt(beta) = 2#

And the circle equation in standard form is
# (x - 3/2)^2 + (y - 2)^2 = 9/4=(3/2)^2 #