The Sphere

For $x^2+y^2+z^2+2ux+2vy+2wz+d=0$: centre $= (-u,-v,-w)$, radius $= \sqrt{u^2+v^2+w^2-d}$. Always ensure the coefficients of $x^2, y^2, z^2$ are each 1 (divide through if needed) before reading $u, v, w, d$.

With diameter endpoints $A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$, the sphere is $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) + (z-z_1)(z-z_2) = 0$. The centre is the midpoint of $AB$ and the radius is half of $AB$.

**Centre $(a,b,c)$, radius $r$:** $(x-a)^2 + (y-b)^2 + (z-c)^2 = r^2$. Concentric spheres share a centre, so they share $u, v, w$; only $d$ differs. To make it pass through a given point, substitute that point to solve for $d$ (passing through the origin forces $d = 0$).

Let $p$ be the perpendicular distance from the sphere's centre to the plane and $r$ the radius:

• $p > r$: no intersection.
• $p = r$: the plane is tangent (touches at one point) — the key NDA condition.
• $p < r$: the plane cuts a circle of radius $\sqrt{r^2 - p^2}$.

A sphere of radius $r$ touches all three axes when its centre is equidistant from each, with each distance equal to $r$. Writing the centre as $(\pm a, \pm a, \pm a)$, the distance from the centre to (say) the $z$-axis is $\sqrt{a^2 + a^2} = a\sqrt2$; setting $a\sqrt2 = r$ gives $a = \tfrac{r}{\sqrt2}$. Each independent sign choice gives a distinct sphere, so there are several such spheres (one per octant configuration).

Set the moving point $(x, y, z)$ equal to the expression you're tracking (centre, centroid, etc.), express the free parameters from those equations, and substitute into the governing condition (the sphere passes through the origin, the plane through a fixed point, etc.). The resulting equation in $x, y, z$ — usually itself a sphere or plane — is the locus.