What is the equation of a sphere in standard form?

The answer is: ##x^2+y^2+z^2+ax+by+cz+d=0##
This is because the sphere is the locus of all
points ##P(xyz)## in the space whose distance from ##C(x_cy_cz_c)## is equal to r.
So we can use the formula of distance from ##P## to ##C## that says:
##sqrt((x-x_c)^2+(y-y_c)^2+(z-z_c)^2)=r## and so:
##(x-x_c)^2+(y-y_c)^2+(z-z_c)^2=r^2##
##x^2+2(x)(x_c) + x_c^2+y^2+2(y)(y_c)+y_c^2+z^2+2(z)(z_c)+z_c^2=r^2##
##x^2+y^2+z^2+ax+by+cz+d=0##
in which
##a=2x_c##;
##b=2y_c##;
##c=2z_c##;
##d=x_c^2+y_c^2+z_c^2-r^2##;
So:
##C(-a/2-b/2-c/2)##
and ##r## if it exists is:
##r=sqrt(x_c^2+y_c^2+z_c^2-d)##.
If the center is in the Origin than the equation is:
##x^2+y^2+z^2=r^2##