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

1 Answer

Answer:

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...