How do you graph the circle x^2+y^2+3x+4y+4=0x2+y2+3x+4y+4=0?

1 Answer
Yonas Yohannes
Jan 25, 2016

(x - 3/2)^2 + (y - 2)^2 = 9/4 (x−32)2+(y−2)2=94
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 x2+y2+3x+4y+4=0

Step 1. Group x and y terms together.
(x^2 + 3x + alpha) + (y^2 + 4y + beta) = -4 + alpha + beta (x2+3x+α)+(y2+4y+β)=−4+α+β
Now the goal to build a perfect by setting
alpha = (3/2)^2 = 9/4; beta = (4/2)^2 = 4α=(32)2=94;β=(42)2=4

Now the circle equation in standard form is:
(x-h)^2 + (y-k)^2 = r^2 (x−h)2+(y−k)2=r2
h = sqrt(alpha) = 3/2; k = sqrt(beta) = 2h=√α=32;k=√β=2

And the circle equation in standard form is
(x - 3/2)^2 + (y - 2)^2 = 9/4=(3/2)^2 (x−32)2+(y−2)2=94=(32)2

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