Question

A plane passes through a fixed point $(p,q,r)$ and cuts the axes in A,B,C, Locus of the centre of the sphere OABC is

• A. $\frac{p}{x}+\frac{q}{y}+\frac{r}{z}=1$
• B. $\frac{p}{x}+\frac{q}{y}+\frac{r}{z}=2$
• C. $\frac{x}{p}+\frac{y}{q}+\frac{z}{r}=1$
• D. $\frac{x}{p}+\frac{y}{q}+\frac{z}{r}=2$

Answer 1

The correct option is B

$\frac{p}{x}+\frac{q}{y}+\frac{r}{z}=2$

Equation of plane $=ax+by+cz=1$ .......$\left(1\right)$
Coordinates of points of plane intersection with (x,y,z) axes are $(1/a,0,0),(0,1/b,0)and(0,0,1/c)$
Let P(l,m,n) be the point of sphere.
${PA}^2={PB}^2={PC}^2={OP}^2$
$({l-\frac{1}{a})}^2+m^2+n^2=l^2+m^2+n^2\Rightarrow l=\frac{1}{2a}$
$\Rightarrow m=\frac{1}{2b}andn=\frac{1}{2c}$ ........$\left(2\right)$
Substitute $\left(2\right)$ in $\left(1\right)$
$\frac{p}{x}+\frac{q}{y}+\frac{r}{z}=2$