analytic geometrygeometrylinear algebra
Write he equation of the sphere that passes through points
$$a(-5,4,1),b(3,4,-5),c(0,0,4),d(0,0,0)$$
I tried to use four points to draw a geometric shape and then calculate the center of this shape on the basis of the circle that passing on four points. But I did not succeed
Here is the book answer
$$x^2+y^2+z^2+54x−58y+4z=0$$
It's a cool game idea. However, the map you're looking for cannot be called a stereographic projection. Stereographic projection projects a sphere onto a whole infinite plane or, in this case, infinite 3-dimensional space $R^3$.
You want to compactify the 3-space to a 3-ball. However, that's clearly not possible if you want to preserve the angles - which is what the stereographic projection does. A 3-sphere is simply not conformally equivalent to a 3-ball because the latter has a boundary.
Imagine that you have such a map $t(P)$ and you are thinking what points $M$ of the 3-sphere get mapped onto - or near - the boundary of the 3-ball, the 2-sphere. Clearly, $M$ must be some singular manifold inside the $S^3$ which is diffeomorphic to an $S^2$. The points of the 3-sphere on the opposite side of the $M$ cannot be found in the 3-ball, as long as the map is one-to-one.
What you could do - if you want to preserve the angles - would be to divide the $S^3$ into the upper and lower hemisphere, $w>0$ or $w<0$, for example, and these two hemispheres are then very close to and naturally mapped to a 3-ball. In this case, you may use the standard stereographic projection and because of the removal of the wrong hemisphere, the infinite 3-plane will shrink to a 3-ball. But you would get two such 3-balls, not one.
Alternatively, you may sacrifice the conservation of the angles and you may "compactify" the 3-plane from the normal stereographic projection by mapping (using polar coordinates) $$(r,\theta,\phi) \to (\arctan r,\theta,\phi).$$ I used arctan because it maps $(0,\infty)$ to a finite interval; $\tanh r$ would do the job, too. And there are others (and simpler) such as $r/(1+r)$. As expected, the angles will be severely distorted near the boundary of the 3-ball which will correspond to the vicinity of the removed pole of the 3-sphere.
Let $P$ and $Q$ be, respectively, the orthocenter and circumcenter of $\triangle ABC$. Let $R$ be the intersection of the perpendicular bisector of $\overline{AB}$ with the altitude from $B$; and let $S$ be the intersection of the perpendicular bisector of ${AC}$ with the altitude from $C$.
Introduce $B^\prime = \overleftrightarrow{AB}\cap\overleftrightarrow{QS}$ and $C^\prime = \overleftrightarrow{AC}\cap\overleftrightarrow{QR}$. Then $\triangle ABC^\prime$ and $\triangle AB^\prime C$ are equilateral.
Observe that $R$ is orthocenter, circumcenter, and incenter for $\triangle ABC^\prime$; likewise, $S$ for $\triangle AB^\prime C$. Therefore, $R$ and $S$ lie on the bisector of $\angle A$. A little angle-chasing shows that $\triangle PRS$ and $\triangle QRS$ are equiangular, hence equilateral, so that $\square PSQR$ is a rhombus, and the result follows.
Best Answer
$$\left | \begin{matrix} x^2 + y^2 + z^2 & x & y & z & 1 \\ x_1^2 + y_1^2 + z_1^2 & x_1 & y_1 & z_1 & 1 \\ x^2_2 + y_2^2 + z_2^2 & x_2 & y_2 & z_2 & 1 \\ x_3^2 + y_3^2 + z_3^2 & x_3 & y_3 & z_3 & 1 \\ x_4^2 + y_4^2 + z_4^2 & x_4 & y_4 & z_4 & 1 \end{matrix} \right | = 0 $$
i.e. $-128(x^2+y^2+z^2-4x-\frac{29}{2}y-4z)=0$