Question

A variable plane passes through a fixed point $(a,b,c)$ and cuts the axes in $A,B$ and $C$ respectively. The locus of the centre of the sphere $OABC,O$ being the origin, is

• A. $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$
• B. $\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=1$
• C. $\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=2$
• D. $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=2$

Answer 1

The correct option is C $\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=2$

Let the points on the coordinate axes be

$A(x_A,0,0)$

$B(0,y_B,0)$

$C(0,0,z_C)$.

Then the equation of the plane which passes through these points, given the x-, y- and z-intercepts, is given as:

$\Rightarrow \frac{x}{x_A}+\frac{y}{y_B}+\frac{z}{z_C}=1$

The fixed point (a,b,c) lies on the plane. Hence,

$\Rightarrow \frac{a}{x_A}+\frac{b}{y_B}+\frac{c}{z_C}=1$

Let the centre of the sphere OABC be given by (X,Y,Z).

Then, the equation of the circle, with radius r:

$\Rightarrow (x-X{)}^2+(y-Y{)}^2+(z-Z{)}^2=r^2$

Since $O(0,0,0)$ lies on the sphere

$\Rightarrow (0-X{)}^2+(0-Y{)}^2+(0-Z{)}^2=r^2$

i.e

$\Rightarrow X^2+Y^2+Z^2=r^2$

A, B, C also lie on the sphere. Hence,

$\Rightarrow (x_A-X{)}^2+(0-Y{)}^2+(0-Z{)}^2=r^2$

$\Rightarrow (0-X{)}^2+(y_B-Y{)}^2+(0-Z{)}^2=r^2$

$\Rightarrow (0-X{)}^2+(0-Y{)}^2+(z_C-Z{)}^2=r^2$

Comparing the 4 equations above:

$\Rightarrow x_A=2X$

$\Rightarrow y_B=2Y$

$\Rightarrow z_C=2Z$

It was already shown that

$\Rightarrow \frac{a}{x_A}+\frac{b}{y_B}+\frac{c}{z_C}=1$

$\therefore \frac{a}{2X}+\frac{b}{2Y}+\frac{c}{2Z}=1$

$\Rightarrow \frac{a}{X}+\frac{b}{Y}+\frac{c}{Z}=2$

Hence, the locus of the centres of the sphere (X,Y,Z) is given by:

$\Rightarrow \frac{a}{x}+\frac{b}{y}+\frac{c}{z}=2$