Answer:
[tex](x-1)^2+ (y+1)^2=3[/tex]
Step-by-step explanation:
[tex]x^2 + y^2 - 2x + 2y - 1 = 0[/tex]
Standard form of the equation is [tex](x-h)^2 + (y-k)^2= r^2[/tex]
To get standard form we apply completing the square method
[tex]x^2-2x+ y^2+ 2y - 1 = 0[/tex]
Take coefficient of x and y . Divide it by 2 and then square it
[tex]\frac{2}{2} =1[/tex] and 1^2=1
Add and subtract 1
[tex](x^2-2x)+(y^2+ 2y) - 1 = 0[/tex]
[tex](x^2-2x+1-1)+(y^2+ 2y+1-1) - 1 = 0[/tex]
[tex](x^2-2x+1)+(y^2+ 2y+1)-1-1- 1 = 0[/tex]
[tex](x^2-2x+1)+(y^2+ 2y+1)-3= 0[/tex]
Now write the parenthesis in square form
[tex](x-1)(x-1)+ (y+1)(y+1)-3= 0[/tex]
[tex](x-1)^2+ (y+1)^2-3= 0[/tex] , add 3 on both sides
[tex](x-1)^2+ (y+1)^2=3[/tex] is the standard form