Question

What is the center of the circle by the equation 3x^2 +12x + 3y^2 - 6y - 33 = 0

Answer 1

Hi,

Explanation:

Method 1:

`3x^2 +12x + 3y^2 - 6y - 33 = 0\\x^2+4x+y^2-2y-11=0\\x^2+4x+4+y^2-2y+1-11-4-1=0\\(x+2)^2+(y-1)^2=4^2\\center\ is\ (-2,1)\ and\ radius=4\\`

Method 2

`f(x,y)=3x^2 +12x + 3y^2 - 6y - 33 = 0\\\\\\\frac{\partial {f}} {\partial{x}} =6x+12=0\Longrightarrow\ 6x=-12 \Longrightarrow\ x=-2\\\\\frac{\partial {f}} {\partial{y}} =6y-6=0\Longrightarrow\ y=1\\\\center\ is\ (-2,1)\\`