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