Question 139830
{{{2x^2+2y^2-8x+16y-32=0}}} Start with the given equation



{{{2x^2-8x+2y^2+16y-32=0}}} Rearrange the terms



{{{2x^2-8x+2y^2+16y=+32}}} Add {{{32}}} to both sides



{{{2(x-2)^2-8+2y^2+16y=+32}}} Complete the square for the x terms



{{{2(x-2)^2-8+2(y+4)^2-32=+32}}} Complete the square for the y terms



{{{2(x-2)^2+2(y+4)^2-40=+32}}} Combine like terms



{{{2(x-2)^2+2(y+4)^2=+32+40}}} Add {{{40}}} to both sides



{{{2(x-2)^2+2(y+4)^2=72}}} Combine like terms



{{{2((x-2)^2+(y+4)^2)=72}}} Factor out the common term 2



{{{(x-2)^2+(y+4)^2=36}}} Divide both sides by 2




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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: (2,-4)


Radius: {{{r=sqrt(36)=6}}}




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Answer: 


So the radius of the circle is 6 units