Question

How do you graph `x^2 + 2x + y^2 + 6y + 6 = 0`?

Answer 1

`x^2+2x+y^2+6y+6=0`

Gather up the terms in `x`, and the terms in `y` and complete the square by combing half the `x` and `y` coefficient for both as follows;

`:. (x+2/2)^2-(2/2)^2 +(y+6/2)^2 - (6/2)^2 +6=0`
`:. (x+1)^2-(1)^2 +(y+3)^2 - (3)^2 +6=0`
`:. (x+1)^2-1 +(y+3)^2 - 9 +6=0`
`:. (x+1)^2 +(y+3)^2 =4`
`:. (x+1)^2 +(y+3)^2 =2^2`

A circle of centre `(a,b)` and radius `r` has equation

`(x-a)^2 +(y-b)^2 =r^2`

Comparing with the above equation we can see that

`(x+1)^2 +(y+3)^2 =2^2`

represents a circle of centre (-1,-3) and radius 2

graph{(x+1)^2 +(y+3)^2 =2^2 [-11.17, 8.83, -7.3, 2.7]}