complete the square to write the equation of the sphere in standard form. find the center and the radius...
x^2+y^2+z^2-4x-6y+4=0
i dont know how to complete the square like this...can anyone help????
Equation of a sphere is given as: $\displaystyle (x-a)^2 + (y-b)^2 + (z-c)^2 = r^2$
where (a,b,c) is the centre of the sphere.
$\displaystyle {\color{red}x^{2}} + {\color{blue}y^2} + {\color{magenta}z^2} - {\color{red}4x} - {\color{blue}6y} + 4 = 0$
$\displaystyle {\color{red}x^{2} - 4x} + {\color{blue}y^{2} - 6y} + {\color{magenta}z^{2}} = -4$
Now complete the square in respect to each variable:
$\displaystyle \left[x^{2} - 4x + {\color{red}\left(\frac{4}{2}\right)^{2}}\right] + \left[y^{2} - 6y + {\color{blue}\left(\frac{6}{2}\right)^{2}}\right] + z^{2} = -4 + {\color{red}\left(\frac{4}{2}\right)^{2}} + {\color{blue}\left(\frac{6}{2}\right)^{2}}$
etc. etc.
We need to group together all x,y, and z's
$\displaystyle (x^2-4x \\\ \\\ )+(y^2-6y \\\ \\\ ) + z^2=-4$
Now we are going to add the square of half the linear coeffient to each side of the equation
$\displaystyle (x^2-4x+4)+(y^2-6y+9)+z^2=-4+4+9$
now we can factor to get
$\displaystyle (x-2)^2+(y-3)^2+z^2=3^2$
So we have a shpere with radius 3 centered at (2,3,0)
Good luck.