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.
Best Answer
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).
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" $OPQR$].