Question 122945

{{{1x^2+2x+1y^2+4y-9=0}}} Start with the given equation



{{{1x^2+2x+1y^2+4y=+9}}} Add {{{9}}} to both sides



{{{1(x+1)^2-1+1y^2+4y=+9}}} Complete the square for the x terms



{{{1(x+1)^2-1+1(y+2)^2-4=+9}}} Complete the square for the y terms



{{{1(x+1)^2+1(y+2)^2-5=+9}}} Combine like terms



{{{1(x+1)^2+1(y+2)^2=+9+5}}} Add {{{5}}} to both sides



{{{1(x+1)^2+1(y+2)^2=14}}} Combine like terms




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Notice how the equation is now in the form {{{(x-h)^2+(y-k)^2=r^2}}}. This means that this conic section is a circle where (h,k) is the center and {{{r}}} is the radius.

So the circle has these properties:


CENTER: ({{{-1}}},{{{-2}}})


Radius: {{{r=sqrt(14)}}}