Question 162030
{{{x^2 + 2x + y^2 + 6y = 15}}} Start with the given equation



{{{x^2 + 2x +highlight(1)+y^2 + 6y = 15+highlight(1)}}} Take half of the "x" coefficient to get {{{(1/2)(2)=1}}}. Now square 1 to get 1. Add this value to both sides (on the left, add it right after the "x" term)




{{{x^2 + 2x +1+ y^2+6y+highlight(9)= 15+1+highlight(9)}}} Take half of the "y" coefficient to get {{{(1/2)(6)=3}}}. Now square 3 to get 9. Add this value to both sides (on the left, add it right after the "y" term)



{{{x^2 + 2x +1+ y^2 + 6y +9= 25}}} Combine like terms.



{{{(x^2 + 2x +1)+ (y^2 + 6y +9)= 25}}} Group the terms into two groups (with three terms in each group)



{{{(x+1)^2+ (y^2 + 6y +9)= 25}}} Factor {{{x^2 + 2x +1}}} to get {{{(x+1)^2}}}



{{{(x+1)^2+ (y+3)^2= 25}}} Factor {{{y^2 + 6y +9}}} to get {{{(y+3)^2}}}



{{{(x-(-1))^2+(y-(-3))^2= 25}}} Rewrite {{{x+1}}} as {{{x-(-1)}}}. Rewrite {{{y+3}}} as {{{y-(-3)}}}



{{{(x-(-1))^2+ (y-(-3))^2= 5^2}}} Replace 25 with {{{5^2}}}



Notice how the equation now fits the form {{{(x-h)^2+(y-k)^2=r^2}}} where in this case {{{h=-1}}}, {{{k=-3}}} and {{{r=5}}}. So this shows us that {{{x^2 + 2x + y^2 + 6y = 15}}} is a circle. So the answer is A