Can someone explain how to do question 3?

2x^2+2y^2-4x+12y+6=0
you want it in the form
(x-a)^2+(y-b)^2=r^2 where (a,b) is the center and r is the radius hence the name standard center radius form.
dividing by 2 and rearranging
x^2-2x+y^2+6y=-3
by completing the square
x^2-2x+1+y^2+6y+9=-3+9+1
(x-1)^2+(y+3)^2=7
this is the standard radius center form (part b)
to get part a we multipy the equation by 2 to get
2(x-1)^2+2(y+3)^2=14 which is choice (2)

Idon’t know i’m in the same boat