A sphere cuts the coordinate axes at points P,Q,R and also passes through the origin. Then which of the equations is satisfied?

Question

A variable plane passes through a fixed point $(a,b,c)$ and cuts the coordinate axes at $P,Q,R$. Then the coordinates of the centre of the sphere passing through$P,Q,R$ and the origin satisfies the equation→

(a)$a/x+b/y+c/z=2$

(b)$x/a+y/b+z/c=3$

(c)$ax+by+cz=1$

(d)$ax+by+cz=a^2+b^2+c^2 $

I know there are ways to do it, but I have thought of an unique geometrical approach.. The plane cuts the axes and makes a tetrahedron.. now as the sphere passes through$P,Q,R$ and the origin, the sphere basically circumscribes the tetrahedron. We can write the equation of the plane as $\frac{x}a +\frac{x}b + \frac{x}c =1$ and therefore can say that it cuts the axes at lengths a, b, and c respectively.. now if we can just find the centre of the tetrahedron, we will get the centre of the sphere.. This is where exactly I am stuck.. I cannot figure out the centre of the tetrahedron. I cannot visualise the structure well.. a detailed answer, with the required geometric figure and proper explanation for the same will be very much helpful. Thank you.

Answer 1

It is equation (a) that is fulfilled.

Edit: Here is a simpler way to reach the result (the first draft is given as an appendix) wjth very few computations.

The variable plane passing through fixed point $F=(a,b,c)$ intersects the coordinate axes in $P=(p,0,0)$, $Q=(0,q,0)$ and $R=(0,0,r)$.

(we assume $p,q,r \ne 0$).

The equation of the plane $PQR$ is known to be :

$$\dfrac xp+\dfrac yq+\dfrac zr=1$$

As $F$ belongs to this plane, wh have :

$$\dfrac ap+\dfrac bq+\dfrac cr=1 \tag{1}$$

Let us now consider the box (parallelepiped) with vertices $O,P,Q,R,O',P',Q',R'$ where $O'=(p,q,r)$. The circumscribed sphere to this box, centered in

$$C=(x_0, y_0, z_0)=(\tfrac p2, \tfrac q2, \tfrac r2) \tag{2}$$

is clearly as well circumscribed to trirectangular tetrahedron $OPQR$, due to central symmetry with respect to $C$ ($P',Q',R'$ being the symmetrical points of $P,Q,R$ resp. with respect to $C$).

If we replace in (1) $(x,y,z)$ by $(x_0, y_0, z_0)$ defined by (2), we get relationship (a).

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Appendix : first version : with the same notations as before.

Points $F,P,Q,R$ being coplanar, the following determinant is zero :

$$\begin{vmatrix}p& 0& 0 &a\\ 0 &q& 0& b\\ 0 &0& r& c\\ 1 &1 &1&1\end{vmatrix}=0 \ \ \iff \ \ \dfrac ap+\dfrac bq+\dfrac cr=1 \tag{1}$$

(this is a classical coplanarity condition : see here).

Besides, the general equation of a sphere passing through the origin with center $(x_0, y_0, z_0)$ is :

$$x^2+y^2+z^2-2 x_0 x - 2 y_0 y -2 z_0 z=0$$

Let us express that this sphere passes through $P, Q, R$ resp.:

$$\begin{cases}p^2 - 2 x_0 p &=&0\\ q^2 - 2 y_0 q &=&0\\ r^2 - 2 z_0 r &=&0 \end{cases}$$

Therefore the coordinates of the center of the sphere are

$$(x_0, y_0, z_0)=(\tfrac p2, \tfrac q2, \tfrac r2)$$

which verify, taking account of (1) :

$$\dfrac{a}{2 x_0}+\dfrac{b}{2 y_0}+\dfrac{c}{2 z_0}=1$$

which is another way to write relationship (a).

Important remark : The center of the sphere has the remarkable geometric property to be the center of the box (= parallelepiped) defined by points $O,P,Q,R$ and $S=(p,q,r)$, [which is not in the base plane $PQR$ of "trirectangular tetrahedron" $OPQR$].

Answer 2

The tetrahedron is not regular, and its center is not the center of the sphere.

Let $O = (0,0,0),$ $P = (p,0,0),$ $Q = (0,q,0),$ and $R=(0,0,r).$ Then the sphere through $O,$ $P,$ and $Q$ intersects the $x,y$ plane in the circle that circumscribes the right triangle $\triangle OPQ.$ The center of the sphere is on the line through the center of that circle perpendicular to the plane of the circle, so once you find the circle you immediately have two of the coordinates of the center of the sphere in terms of $p,$ $q,$ and $r.$ Now look at how the sphere intersects either the $x,z$ plane or the $y,z$ plane and you will just as quickly find the third coordinate.

Then you will have all three coordinates $(x,y,z)$ of the sphere's center in terms of $p,$ $q,$ and $r.$

You can also write an equation that $(a,b,c)$ must satisfy since that point is on the plane through $(p,0,0),$ $(0,q,0),$ and $(0,0,r).$

Now use what you know about $p,$ $q,$ and $r$ to get an equation that involves only $a,$ $b,$ $c,$ $x,$ $y,$ and $z.$