Given the following equation:
[tex]x^2+y^2+8x-12y+120=144[/tex]We know that the equation of the circle with radius r and center (h,k) is:
[tex](x-h)^2+(y-k)^2=r^2[/tex]in this case, we can associate the terms with x and y separately to get the following:
[tex]\begin{gathered} x^2+y^2+8x-12y+120=144 \\ \Rightarrow(x^2+8x)+(y^2-12y)=144-120=24 \\ \Rightarrow(x^2+8x)+(y^2-12y)=24 \end{gathered}[/tex]next, we can complete the square on both summands to get the following:
[tex](x^2+8x+16)+(y^2-12y+36)=24+16+36[/tex]notice that we must add the terms to complete the square on both sides of the equation! IF we factorize the polynomials we get:
[tex]\begin{gathered} (x^2+8x+16)+(y^2-12y+36)=24+16+36_{} \\ \Rightarrow(x+4)^2+(y-6)^2=76 \end{gathered}[/tex]therefore, the x-coordinate of the center is h=-4
the y-coordinate of the ceter is k = 6
the radius is r = sqrt(76)