Question 140867

{{{x^2+8x+y^2+6y-24=0}}} Start with the given equation



{{{x^2+8x+y^2+6y=+24}}} Add {{{24}}} to both sides



{{{(x+4)^2-16+y^2+6y=+24}}} Complete the square for the x terms



{{{(x+4)^2-16+(y+3)^2-9=+24}}} Complete the square for the y terms



{{{(x+4)^2+(y+3)^2-25=+24}}} Combine like terms



{{{(x+4)^2+(y+3)^2=+24+25}}} Add {{{25}}} to both sides



{{{(x+4)^2+(y+3)^2=49}}} 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: (-4,-3)


Radius: {{{r=sqrt(49)=7}}}