Answer :
Completing the square consists of coverting quadratic equations with the following form
[tex]ax^2+bx+c[/tex]into a squared binomial plus a constant:
[tex]ax^2+bx+c=a(x-h)^2+k[/tex]When we expand a squared binomial, we have:
[tex](a\pm b)^2=a^2\pm2ab+b^2[/tex]We want to rewrite the following expression using this method:
[tex]x^2-4x+y^2+8y=-4[/tex]We can write a binomial for x and another for the y variable.
Let's start by rewriting the expression for x. The terms in our expression that have the x-variable as a factor are:
[tex]x^2-4x[/tex]If we compare this part of the equation with the expanded binomial form, we can assume the linear term is the result of the "cross product".
[tex]\begin{gathered} -2ab=-4x \\ a=x\implies b=2 \end{gathered}[/tex]The missing term is the square of 2. If we add and subtract the square of 2 to our expression, it will mantain its original value, allowing us to rewrite the expression on the desired form:
[tex]\begin{gathered} x^2-4x=x^2-4x+2^2-2^2 \\ =x^2-4x+4-4 \\ =(x^2-4x+4)-4 \\ =(x-2)^2-4 \\ \\ \implies x^2-4x=(x-2)^2-4 \end{gathered}[/tex]If we substitute this binomial on the original equation, we have:
[tex]\begin{gathered} x^{2}-4x+y^{2}+8y=-4 \\ (x^2-4x)+y^2+8y=-4 \\ ((x-2)^2-4)+y^2+8y=-4 \\ (x-2)^2-4+y^2+8y=-4 \\ (x-2)^2+y^2+8y=0 \end{gathered}[/tex]Repeating the process for the terms with the y-variable, we have:
[tex]\begin{gathered} y^2+8y=y^2+8y+16-16 \\ =y^2+2(4)(y)+4^2-16 \\ =(y+4)^2-16 \end{gathered}[/tex]Then, our circle equation is:
[tex]\begin{gathered} (x-2)^{2}+y^{2}+8y=0 \\ (x-2)^2+(y^2+8y)=0 \\ (x-2)^2+(y+4)^2-16=0 \\ (x-2)^2+(y+4)^2=16 \end{gathered}[/tex]The standard circle equation has the following form:
[tex](x-h)^2+(y-k)^2=r^2[/tex]where (h, k) are the coordinates of the center and r represents the radius.
If we compare our circle equation with this form, our circle is centered at (2, -4) and has a radius equals to 4.