Answer :
The standard equation of a circle can be wriiten as;
[tex](x-h)^2+(y-k)^2=r^2[/tex]So, to get the value of h,k and r we need to express the given equation in this form.
The equation is given as;
[tex]x^2+y^2-6x+2y+9=0[/tex]Then we will try to express this equation in the standard equation of circle form.
[tex]x^2-6x+9+y^2+2y+1=1[/tex]Note that we added 1 to both sides so that we can factorise the equation.
Factorizing, we have;
[tex](x-3)^2+(y+1)^2=1[/tex]Then finally we have;
[tex](x-3)^2+(y-(-1))^2=1^2[/tex]Comparing the equation to the standard form we have;
[tex]\begin{gathered} h=3 \\ k=-1 \\ radius\rightarrow r=1 \end{gathered}[/tex]Then lastly we need to graph the equation;
From the graph, we can see that the only intercept we have is at;
[tex]\begin{gathered} x=3 \\ \text{ point }\rightarrow(3,0) \end{gathered}[/tex]There is no intercept on the y axis.
