Raj tested his new flashlight by shinning it on his bedroom wall the beam of the light can be described by the equation x^2 y^2-2x-4y-31=0 how many inches wide is the beam of light on the wall

I can get it into standard form of that certain conic section by completing the square

(x^2-2x)+(y^2-4y)-31=0 (x^2-2x+1-1)+(y^2-4y+4-4)-31=0 (x-1)^2-1+(y-2)^2-4-31=0 (x-1)^2+(y-2)^2-36=0 (x-1)^2+(y-2)^2=36 (x-1)^2+(y-2)^2=6^2

in form
(x-h)^2+(y-k)^2=r^2
r=radius
so how wide it is is the diameter which is 2 times of radius

6=radius
6 times 2=12

12 inches I assume if the units in equation is in inches

Intinisbinidbifbisbe