The standard equation of a circle is [tex]\displaystyle{ (x-a)^2+(y-b)^2=r^2[/tex],
where (a, b) is the center of the circle, and r is the radius.
So, if we write [tex]x^2 - 2x + y^2 - 8y + 1 = 0[/tex] in the above form, the center and the radius will be obvious to us.
Note that [tex]x^2-2x[/tex] is very close to [tex](x-1)^2=x^2-2x+1[/tex] except for the 1.
Similarly, we notice that [tex]y^2-8y[/tex] is like [tex](y-4)^2=x^2-8x+16[/tex], without the 16.
So,
[tex]x^2 - 2x + y^2 - 8y + 1 = 0\\\\(x^2 - 2x) + (y^2 - 8y) + 1 = 0\\\\(x^2 - 2x+1)-1 + (y^2 - 8y+16)-16 + 1 = 0\\\\(x-1)^2+(y-4)^2-16=0\\\\(x-1)^2+(y-4)^2=16=4^2[/tex].
Answer:
[tex](x-1)^2+(y-4)^2=4^2[/tex]
