Answer
(a) The circle has a center (3,-1) and a radius of 1 unit long.
(b) See below.
(c) The only x-intercept is $(3,0)$. There are no y-intercepts.
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:
$x^2+y^2-6x+2y+9=0$
$x^2-6x+y^2+2y=-9$
$x^2-6x+(\frac{6}{2})^2+y^2+2y+(\frac{2}{2})^2=-9+(\frac{6}{2})^2+(\frac{2}{2})^2$
$(x-3)^2+(y+1)^2=-9+9+1$
$(x-3)^2+(y+1)^2=1$
$(x-3)^2+(y+1)^2=1^2$
Now we can find the center (h,k) and the radius.
The x-intercepts are all points of a graph when y=0:
$(x-3)^2+(0+1)^2=1$
$(x-3)^2+1=1$
$(x-3)^2=0$
$\sqrt{(x-3)^2}=\sqrt0$
There is only one x-intercept:
$x-3=0\rightarrow x=3$
The y-intercepts are all points of a graph when x=0:
$(0-3)^2+(y+1)^2=1$
$9+(y+1)^2=1$
$(y+1)^2=-8$
$\sqrt{(y+1)^2}=\sqrt{-8}$
Taking the square root of a negative number is not possible; therefore, the graph doesn't intercept the y-axis at any point.
