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Sagot :
the solution is
[tex](x^{}-2)^2+(y+4)^2=16[/tex]the center is in (x,y)=(2,4)
the radius is 4
to solve this, first we agroup the x's and y's,
[tex](x^2-4x)+(y^2+8y)=-4[/tex]normaly, we would need to divide the whole equation by the coeficient of the squares
in this case the coeficient is 1 so we don't need to do that
now, we need to look at the coeficient of the x and y (not squared)
for x is -4 and for y is +8
we divide them bi two, and square them
-4/2=-2 => -2^2=4
8/2=4 => 4^2=16
and now we add this terms in the parentheses and add the to the right side of the equation
[tex](x^2-4x+4)+(y^2+8y+16)=-4+4+16[/tex]now, we need to simplify the parentheses and let them expresed as binomial squares
[tex](x-2)^2+(y+4)^2=16[/tex]and thats the equation. now the term inside the parentheses that aren't x or y, tell us the center.
center (x,y)=(2,-4)
and the radius is the square root of the term in the right side
[tex]\text{radius: }\sqrt[]{16}=4[/tex]
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