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 $AB$. The form says $\vec{PA}\cdot\vec{PB} = 0$ for every point $P$ of the sphere — the angle in a semicircle is a right angle — so by Pythagoras $PA^2 + PB^2 = AB^2$ for every $P$ on the sphere; a locus question "$PA^2+PB^2 = k$" with $k = AB^2$ IS this sphere, and with other $k$ it is a concentric one.

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 $d = 0$). The smallest sphere through three given points has those points on a GREAT circle: its centre is the circumcentre of the triangle they form and its radius the circumradius $R = \dfrac{abc}{4\Delta}$. For an equilateral triangle of side $a$, $R = \dfrac{a}{\sqrt3}$ and the centre is the centroid — so for $(1,0,0),(0,1,0),(0,0,1)$ (side $\sqrt2$) the smallest sphere has centre $(\tfrac13,\tfrac13,\tfrac13)$ and radius $\sqrt{2/3}$. Every other sphere through the three points is larger, with its centre on the line through the circumcentre perpendicular to the plane.

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). A different configuration — the sphere THROUGH the origin and the three axis intercepts $A(\alpha,0,0)$, $B(0,\beta,0)$, $C(0,0,\gamma)$: substituting the four points into $x^2+y^2+z^2+2ux+2vy+2wz+d=0$ gives $d=0$, $u=-\alpha/2$, $v=-\beta/2$, $w=-\gamma/2$, so the equation is $x^2+y^2+z^2-\alpha x-\beta y-\gamma z=0$ and the centre is $(\alpha/2,\ \beta/2,\ \gamma/2)$ — the midpoint of the box diagonal $O$ to $(\alpha,\beta,\gamma)$, radius $\tfrac12\sqrt{\alpha^2+\beta^2+\gamma^2}$. Locus questions follow: if the plane $\tfrac x\alpha+\tfrac y\beta+\tfrac z\gamma=1$ is constrained (e.g. passes through a fixed point $(a,b,c)$, or the intercepts satisfy $\tfrac1{\alpha^2}+\tfrac1{\beta^2}+\tfrac1{\gamma^2}=\tfrac1{p^2}$ for a fixed distance $p$ from $O$), write the centre as $(x,y,z)=(\alpha/2,\beta/2,\gamma/2)$ and substitute $\alpha=2x$ etc.: the locus is $\tfrac a x+\tfrac b y+\tfrac c z=2$, respectively $\tfrac1{x^2}+\tfrac1{y^2}+\tfrac1{z^2}=\tfrac4{p^2}$.

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.