Identify the standard form of a circle equation x^2+2x+y^2-2y=2
Accepted Solution
A:
Answer:The standard from of the expression is: [tex](x +1)^2 + (y-1)^2 = (2)^2[/tex]Step-by-step explanation:Here the given expression is : [tex]x^2+2x+y^2-2y=2[/tex]Now, the standard form of a circle is given as :[tex](x-h)^2 + (y -k)^2 = r^2[/tex]Here, (h,k) = Coordinates of Center, r = RadiusAlso, use the algebraic identity:[tex](a \pm b)^2 = a^2 + b^2 \pm 2ab\\[/tex]Now, converting the given expression in the standard form, we get:[tex]x^2+2x+y^2-2y=2[/tex]Add 2 on both sides of the equation, we get:[tex]x^2+2x+y^2-2y +2=2 + 2\\\implies x^2+2x + 1 + y^2-2y + 1 = 4\\\implies (x^2+2x + 1 )+ (y^2-2y + 1) = 4\\\implies (x +1)^2 + (y-1)^2 = (2)^2[/tex]So, here the standard from of the expression is:[tex](x +1)^2 + (y-1)^2 = (2)^2[/tex]Center coordinates here are (h,k) = ( -1 ,1) and Radius = 2 units