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:

$x^2+y^2+z^2=r^2$ ,