Question

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

Answer 1

This is a circle with center at `C(5, 5)` and with radius`=3`

Explanation:

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

`x^2-10x+y^2-10y=-41`
Add 50 on both sides of the equation
`(x^2-10x+25)+(y^2-10y+25)=-41+50`

`(x-5)^2+(y-5)^2=9`
We have a circle with Center at `(h, k)=(5, 5)`
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...