Question 141450
{{{9x^2-108x+4y^2+8y+184=0}}}
{{{9(x^2-(108/9)x+(108/18)^2)+4(y^2+2y+1)+184-4-9(108/18)^2=0}}}<--complete the squares
{{{9(x-6)^2+4(y+1)^2+184-4-9*36=0}}}<---factor
{{{9(x-6)^2+4(y+1)^2=144}}}<--simplify, shift constants to right-side
{{{(x-6)^2+(4/9)(y+1)^2=144/9}}}<---divide 9 through to make x term standard
{{{(1/4)(x-6)^2+(1/9)(y+1)^2=144/36}}}<---divide 4 through to make the y term standard
{{{(1/(16))(x-6)^2+(1/36)(y+1)^2=1}}}<---simplify it into standard form by dividing 4 through in order to make the right side 1.
{{{(1/(4)^2)(x-6)^2+(1/(6)^2)(y+1)^2=1}}}
It should be evident now that this is an ellipse, not a circle.

The center is (h,k) when in standard form: (6,-1).