Answer:
Step-by-step explanation:
You have:
[tex]x^2+y^2-8x+27=0[/tex]
Since the variable 'x' appears twice, I am going to choose to group together the terms as to create a 2nd degree trinomial:
[tex](x^2-8x+27)+y^2=0[/tex]
This allows us to complete the square and factor the trinomial:
[tex](x^2-8x+16-16+27)+y^2=0[/tex]
[tex](x^2-8x+16)+11+y^2=0[/tex]
[tex](x-4)^2+y^2+11=0[/tex]
[tex](x-4)^2+(y-0)^2=-11[/tex]
This is not a valid circle equation in the real plane. It has a complex radius of:
[tex]r=i\sqrt{11}[/tex]
And is centered at the point:
[tex](4,0)[/tex]
