Question

A variable plane passes through a fixed point $(a,b,c)$ and cuts the coordinate axes at $A,B$ and $C.$ Then the locus of the centre of the sphere $OABC$ is

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

Answer 1

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

Let $(\alpha ,\beta ,\gamma )$ be any point on locus.
So, equation of shpere is given by,
$(x-\alpha {)}^2+(y-\beta {)}^2+(z-\gamma {)}^2=(0-\alpha {)}^2+(0-\beta {)}^2+(0-\gamma {)}^2{x}^2+y^2+z^2-2\alpha x-2\beta y-2\gamma z=0$
for intersection on $x-$axis : put $y=z=0$
$\Rightarrow x^2-2\alpha x=0\Rightarrow x=0,2\alpha$
Thus plane meets $x-$axis at $(0,0,0)$ and $(2\alpha ,0,0)$
similarly on $y-$axis at $(0,0,0)$ and $(0,2\beta ,0)$ and on $z-$axis at $(0,0,0)$ and $(0,0,2\gamma )$
Thus the eqaution of plane through $A,B,C$ is :
$\frac{x}{2\alpha }+\frac{y}{2\beta }+\frac{z}{2\gamma }=1$
Since it passes through, $(a,b,c)$
$\Rightarrow \frac{a}{2\alpha }+\frac{b}{2\beta }+\frac{c}{2\gamma }=1\Rightarrow \frac{a}{\alpha }+\frac{b}{\beta }+\frac{c}{\gamma }=2$
hence, locus of $(\alpha ,\beta ,\gamma )$ is
$\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=2$