The center-radius form of the circle equation
[tex](x-h)^2+(y-k)^2=r^2[/tex]
(h, k) - center
r - radius
We have:
[tex]x^2+y^2-4x-4y+4=0[/tex]
Use
[tex](a\pmb)^2=a^2\pm2ab+b^2\qquad(*)[/tex]
[tex]x^2-4x+y^2-4y+4=0\qquad\text{subtract 4 from both sides}\\\\x^2+2(x)(2)+y^2-2(y)(2)=-4\qquad\text{add two times} \ 2^2\ \text{to both sides}\\\\\underbrace{x^2+2(x)(2)+2^2}_{(*)}+\underbrace{y^2-2(y)(2)+2^2}_{(*)}=2^2+2^2-4\\\\(x+2)^2+(y-2)^2=4+4-4\\\\(x+2)^2+(y-2)^2=4\\\\(x+2)^2+(y-2)^2=2^2\\\\Answer:\ \boxed{radius:\ r=2}[/tex]
Considering the equation of the circle, the length of the radius is of 2 units.
The equation of a circle of center [tex](x_0, y_0)[/tex] and radius r is given by:
[tex](x - x_0)^2 + (y - y_0)^2 = r^2[/tex]
In this problem, the equation is given by:
[tex]x^2 + y^2 - 4x - 4y + 4 = 0[/tex]
Completing the squares, we have that:
[tex]x^2 - 4x + y^2 - 4y = -4[/tex]
[tex](x - 2)^2 + (y - 2)^2 = -4 + 4 + 4[/tex]
[tex](x - 2)^2 + (y - 2)^2 = 4[/tex]
Hence, the radius is given by:
[tex]r^2 = 4[/tex]
[tex]r = 2[/tex]
The length of the radius is of 2 units.
More can be learned about the equation of a circle at https://brainly.com/question/24307696
