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. $\frac{p}{x} + \frac{q}{y} + \frac{r}{z} = 2$
• B. $\frac{p}{x} + \frac{q}{y} + \frac{r}{z} = 1$
• C. $\frac{p}{x} + \frac{q}{y} + \frac{r}{z} = 3$
• D. None of these

Answer 1

Answer: (A)

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

Answer 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 $x ^ { 2 } + y ^ { 2 } + z ^ { 2 } - a x - b y - c z = 0$

If its centre $\left( x _ { 1 } , y _ { 1 } , z _ { 1 } \right)$, then $x _ { 1 } = \frac { a } { 2 } , y _ { 1 } = \frac { b } { 2 } , z _ { 1 } = \frac { c } { 2 }$

∴ $a = 2 x _ { 1 } , b = 2 y _ { 1 } , c = 2 z _ { 1 }$

∴ Locus of centre of sphere $\frac { p } { x } + \frac { q } { y } + \frac { r } { z } = 2$.