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# if i am given an equation for circle how do i find its center

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

Guest Nov 30, 2014

#2
+20585
+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
+94098
+5

\$\$\\4x^2+2y^2=11\\\\
\$This is not a circle. It is an ellipse\$\$\$
.

The centre is (0,0) because it is just x2 and y2

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

http://www.mathopenref.com/coordgeneralellipse.html

Melody  Nov 30, 2014
#2
+20585
+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