Answer
$\frac{(x-2)^2}{9}+\frac{(y+3)^2}{4}=1$
Work Step by Step
$4x^2+9y^2-16x+54y+61=0$
$4x^2-16x+9y^2+54y+61=0$
$4(x^2-4x)+9(y^2+6y)+61=0$
$4(x^2-4x+4)-16+9(y^2+6y+9)-81+61=0$
$4(x-2)^2+9(y+3)^2=16+81-61$
$4(x-2)^2+9(y+3)^2=36$
$\frac{(x-2)^2}{9}+\frac{(y+3)^2}{4}=1$
We can write the general form of an ellipse:
$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2} = 1$
The equation in the question is an equation of an ellipse, where $a = 3, b = 2, h = 2,$ and $k = -3$
