Question

A plane passes through a fixed point (p,q,r) and cut the axes in A,B,C. Then the locus of the centre of the sphere OABC is

• A. $\frac{p}{x}+\frac{q}{y}+\frac{r}{2}=2$
• B. $\frac{p}{x}+\frac{q}{y}+\frac{r}{2}=2$
• C. $\frac{p}{x}+\frac{q}{y}+\frac{r}{2}=3$
• D. $Noneofthese$

Answer 1

The correct option is A

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

Let the co-ordinates of A, B and C be (a,0,0), (0,b,0) and (0,0,c) respectively.
The equation of the plane is $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$
Also, it passes through (p,q,r). So, $\frac{p}{a}+\frac{q}{b}+\frac{z}{c}=1$
Also equation of sphere passes through A,B,C will be $x^2+y^2+z^2-ax-by-ca=0$
If its centre $(x_1,y_1,z_1)$, then $x_1=\frac{a}{2},y_1=\frac{b}{2},z_1=\frac{c}{2}$
$\therefore a=2x_1,b=2y_1,c=2z_1$.
$\therefore$ Locus of centre of sphere $\frac{p}{x}+\frac{q}{y}+\frac{r}{2}=2$.