What is the equation of a sphere in standard form?

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Answer 1 · 2015-02-01T12:56:55

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(x,y,z)# in the space whose distance from #C(x_c,y_c,z_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#,