Answer
Please see the graph.
Work Step by Step
$\frac{(x-2)^2}{4}+ (y-1)^2 =1$
This equation is of an ellipse. This equation was almost for a circle, if it wasn't for the denominator of 4 in the first expression.
The center of the ellipse is at $(2,1)$.
$x=0$
$\frac{(x-2)^2}{4}+ (y-1)^2 =1$
$\frac{(0-2)^2}{4}+ (y-1)^2 =1$
$\frac{(-2)^2}{4}+ (y-1)^2 =1$
$\frac{4}{4}+ (y-1)^2 =1$
$1 + (y-1)^2 =1$
$(y-1)^2=0$
$\sqrt {(y-1)^2} =\sqrt 0$
$y-1=0$
$y-1+1=0+1$
$y=1$
$(0,1)$ is also on the ellipse.
$x=2$
$\frac{(x-2)^2}{4}+ (y-1)^2 =1$
$\frac{(2-2)^2}{4}+ (y-1)^2 =1$
$\frac{(0)^2}{4}+ (y-1)^2 =1$
$\frac{0}{4}+ (y-1)^2 =1$
$0 + (y-1)^2 =1$
$(y-1)^2=1$
$\sqrt {(y-1)^2} =\sqrt 1$
$y-1=±1$
$y-1=1$
$y-1+1=1+1$
$y=2$
$y-1=-1$
$y-1+1=-1+1$
$y=0$
$(2,0)$ and $(2,2)$ are also on the ellipse.
