the equation of the circle is x(to the second power) + y(to the second power) - 4x + 2y =11

Guest Nov 30, 2014

#2**+10 **

**the equation of the circle is x(to the second power) + y(to the second power) - 4x + 2y =11**

**$$x^2+y^2-4x+2y=11\\\\ \underbrace{( x^2 - 4x )}_{=(x-2)^2-4} + \underbrace{(y^2 +2y )}_{=(y+1)^2-1} = 11 \\\\ (x-2)^2 +(y+1)^2 = 11 +1+4\\\\ (x \textcolor[rgb]{1,0,0}{ -2})^2 +(y \textcolor[rgb]{0,0,1}{+1})^2 = 16\\\\ \boxed{ Center_x = -(\textcolor[rgb]{1,0,0}{ -2}) = 2 }\\ \boxed { Center_y = -(\textcolor[rgb]{0,0,1}{ +1}) = -1 }$$**

heureka
Dec 1, 2014

#1**+5 **

$$\\4x^2+2y^2=11\\\\

$This is not a circle. It is an ellipse$$$.

The centre is (0,0) because it is just x^{2} and y^{2}

If, for example, the centre was (3,-5) then it would be (x-3)^{2} and (y+5)^{2}

Here it is :)

https://www.desmos.com/calculator/v35v5nrxpg

Here is an artical explaining elipses

Melody
Nov 30, 2014

#2**+10 **

Best Answer

**the equation of the circle is x(to the second power) + y(to the second power) - 4x + 2y =11**

**$$x^2+y^2-4x+2y=11\\\\ \underbrace{( x^2 - 4x )}_{=(x-2)^2-4} + \underbrace{(y^2 +2y )}_{=(y+1)^2-1} = 11 \\\\ (x-2)^2 +(y+1)^2 = 11 +1+4\\\\ (x \textcolor[rgb]{1,0,0}{ -2})^2 +(y \textcolor[rgb]{0,0,1}{+1})^2 = 16\\\\ \boxed{ Center_x = -(\textcolor[rgb]{1,0,0}{ -2}) = 2 }\\ \boxed { Center_y = -(\textcolor[rgb]{0,0,1}{ +1}) = -1 }$$**

heureka
Dec 1, 2014