How do you graph x^2 + y^2 - 10x - 10y + 41 = 0?

This is a circle with center at $C \left(5 , 5\right)$ and with radius$= 3$

Explanation:

from the given
${x}^{2} + {y}^{2} - 10 x - 10 y + 41 = 0$
Rearrange the terms then use "Complete the square" method

${x}^{2} - 10 x + {y}^{2} - 10 y = - 41$
Add 50 on both sides of the equation
$\left({x}^{2} - 10 x + 25\right) + \left({y}^{2} - 10 y + 25\right) = - 41 + 50$

${\left(x - 5\right)}^{2} + {\left(y - 5\right)}^{2} = 9$
We have a circle with Center at $\left(h , k\right) = \left(5 , 5\right)$
with radius $r = 3$

graph{x^2+y^2-10x-10y+41=0 [-20, 20, -10, 10]}

God bless... I hope the explanation is useful...