# A question related to Plane and Sphere

The problem: A variable plane passes through a fixed point (a,b,c) and cuts the coordinate axes at P, Q, R (where none of P, Q, R is the origin). The co-ordinates (x,y,z) of the center of the sphere passing through P, Q, R and the origin satisfy 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 = a2 + b2 + c2

I took the variable plane as x/p + y/q + z/r = 1 and the sphere to be

(X-x)2 + (Y-y)2 + (Z-z)2 = R2 and substituting the values of (p, 0, 0), (q, 0, 0), (r, 0, 0) & (0, 0, 0) in it, I eventually arrived at:

p2 + q2 + r2 = 2px + 2qy + 2rz, which is an equation of a plane in x, y, z. Hence, I was able to eliminate option A(certainly not a plane).

Any hint from here? Other approaches also welcome.

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Bhaskar, you might want to post your own solution and accept it if you have found it yourself, so that this question is settled. –  Ｊ. Ｍ. May 6 '11 at 3:55
The values should be $(p, 0, 0), (0, q, 0), (0, 0, r)$ as a plane can only cut each axis once. –  Ross Millikan May 25 '11 at 0:27

You have points $P(p,0,0),Q(0,q,0),R(0,0,r)$ on the axes and the origin $O$. Denote by $K$ the circumcenter of the triangle $PQR$. Since all the angles $\angle POQ,\angle QOR,\angle ROP$ are right, the sphere which passes through $P,Q,R,O$ has center $K$. To find $K$ easily, complete the $PQRO$ to a box, and $K$ would be then the midpoint of the great diagonal i.e. $K(p/2,q/2,r/2)$.
The plane has equation $k(x-a)+l(y-b)+m(z-c)=0$, (with variables $k,l,m$) written equivalently $\frac{x}{\frac{ka+lb+mc}{k}}+\frac{y}{\frac{ka+lb+mc}{l}}+\frac{z}{\frac{ka+lb+mc}{m}}=1$. This yields $p=\frac{ka+lb+mc}{k},q=\frac{ka+lb+mc}{l},r=\frac{ka+lb+mc}{m}$, and the coordinates of the center of the sphere are $x=\frac{ka+lb+mc}{2k},y=\frac{ka+lb+mc}{2l},\frac{ka+lb+mc}{2m}$.