Question 1046374

The equation of a circle in standard form is: {{{(x-h)^2+(y-k)^2=r^2}}}

 {{{x^2 + y^2 + 4x - 2y = -1}}}....write it in standard form

 {{{(x^2 + 4x) + (y^2 - 2y) = -1}}}.............complete squares

{{{(x^2 + 4x+b^2) -b^2+ (y^2 - 2y+b^2)-b^2 = -1}}}

{{{(x^2 + 4x+2^2) -2^2+ (y^2 - 2y+1^2)-1^2 = -1}}}

{{{(x+2)^2 -4+ (y - 1)^2-1 = -1}}}

{{{(x+2)^2 + (y - 1)^2-5 = -1}}}

{{{(x+2)^2 + (y - 1)^2 = -1+5}}}

{{{(x+2)^2 + (y - 1)^2 = 4}}}

{{{(x+2)^2 + (y - 1)^2 = 2^2}}}=> {{{r=2}}}

so, your answer is: A) {{{2}}}