Respuesta :
We have the equation:
[tex]x^2-4x+y^2+8y=-4[/tex]And we have to writte it in a way that we can relate it to the equation of a circle. The equation of a circle centered around point (a,b) and with a radius r is given by:
[tex](x-a)^2+(y-b)^2=r^2[/tex]Now let's make some operations in both equations:
[tex]\begin{gathered} x^2-4x+y^2+8y=-4 \\ x^2-4x+y^2+8y+4=0 \end{gathered}[/tex][tex]\begin{gathered} (x-a)^2+(y-b)^2=r^2 \\ (x-a)^2+(y-b)^2-r^2=0 \end{gathered}[/tex]Since both equations are equal to 0 we can merge them into one great equation:
[tex](x-a)^2+(y-b)^2-r^2=x^2-4x+y^2+8y+4[/tex]Now let's work with this one:
[tex]\begin{gathered} (x-a)^2+(y-b)^2-r^2=x^2-4x+y^2+8y+4 \\ x^2-2ax+a^2+y^2-2by+b^2-r^2=x^2-4x+y^2+8y+4 \\ (x^2-2ax)+(y^2-2by)+(a^2+b^2-r^2)=(x^2-4x)+(y^2+8y)+4 \end{gathered}[/tex]Where I divided the terms on each side of the equation in three groups: the ones multiplied by x, the ones multiplied by y and the independent terms not multiplid by x nor y. The key here is that one group in one side of the equation must be equal to its counterpart on the other side:
[tex]\begin{gathered} x^2-2ax=x^2-4x \\ y^2-2by=y^2+8y \\ a^2+b^2-r^2=4 \end{gathered}[/tex]Using these equations we can find the value of a, b and r:
[tex]\begin{gathered} x^2-2ax=x^2-4x \\ 2ax=4x \\ a=\frac{4}{2}=2 \end{gathered}[/tex][tex]\begin{gathered} y^2-2by=y^2+8y \\ -2by=8y \\ b=-\frac{8}{2}=-4 \end{gathered}[/tex]We have a=2 and b=-4 so we can find r using the third equation:
[tex]\begin{gathered} a^2+b^2-r^2=4 \\ 2^2+(-4)^2-r^2=4 \\ 4+16-r^2=4 \\ r^2=16 \\ r=4 \end{gathered}[/tex]In summary, the circle is centered around point (2,-4) and has a radius of 4.
