[tex]\text{The standard form of the circle:}\\\\(x-h)^2+(y-k)^2=r^2\\\\\text{h and k are the x and y coordinates of the circle}[/tex]
[tex]\text{We have}\\\\x^2-4x+y^2+8y-29=0\\\\\text{use:(*)}(a\pm b)^2=a^2\pm2ab+b^2[/tex]
[tex]x^2-2x\cdot2+y^2+2y\cdot4=29\ \ \ |+2^2\ \ |+4^2\\\\\underbrace{x^2-2x\cdot2+2^2}_{(*)}+\underbrace{y^2+2y\cdot4+4^2}_{(*)}=29+2^2+4^2\\\\(x-2)^2+(y+4)^2=49[/tex]
[tex]\text{Answer: the center (2;-4), the radius}\ r=\sqrt{49}=7[/tex]
