Answer
The circle has a center (1,-2) and a radius of 3 units.
There are two x-intercepts: $(-1.24,0) \text{ and } (3.24,0)$.
There are two y-intercepts: $(0,-4.83) \text{ and } (0,0.83)$.
Work Step by Step
First, we need to make sure that the equation is in standard form $(x-h)^2+(y-k)^2=r^2$
In this case, we need to complete the square to achieve the standard form:
$x^2+y^2-2x+4y-4=0$
$x^2-2x+y^2+4y=4$
$x^2-2x+(\frac{2}{2})^2+y^2+4y+(\frac{4}{2})^2=4+(\frac{2}{2})^2+(\frac{4}{2})^2$
$(x-1)^2+(y+2)^2=4+1+4$
$(x-1)^2+(y+2)^2=9$
$(x-1)^2+(y+2)^2=3^2$
Now we can find the center (h,k) and the radius.
The x-intercepts are all points of a graph when y=0:
$(x-1)^2+(0+2)^2=9$
$(x-1)^2+4=9$
$(x-1)^2=5$
$\sqrt{(x-1)^2}=\sqrt5$
There are two x-intercepts:
$x_1-1=-\sqrt8\rightarrow x_1=-\sqrt5+1\approx-1.24$
$x_2-1=\sqrt8\rightarrow x_2=\sqrt5+1\approx3.24$
The y-intercepts are all points of a graph when x=0:
$(0-1)^2+(y+2)^2=9$
$1+(y+2)^2=9$
$(y+2)^2=8$
$\sqrt{(y+2)^2}=\sqrt{8}$
There are two y-intercepts:
$y_1+2=-\sqrt8\rightarrow y_1=-\sqrt8-2\approx-4.83$
$y_2+2=\sqrt8\rightarrow y_2=\sqrt8-2\approx0.83$
