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The equation x2 + y2 − 2x + 2y − 1 = 0 is the general form of the equation of a circle. What is the standard form of the equation?

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x^2 + y^2 - 2x + 2y - 1 = 0

(x^2 - 2x) + (y^2 + 2y) - 1 = 0

(x^2 - 2x + 1) + (y^2 + 2y + 1) - 1 - 1 - 1 = 0

(x - 1)^2 + (y + 1)^2 - 3 = 0

(x - 1)^2 + (y + 1)^2 = 3


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

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