Question

The plane $\frac{x}{2}+\frac{y}{3}+\frac{z}{4}=1$ meets the coordinate axes in $A,B,C$. The equation of sphere through $O,A,B,C$ is

• A. $x^2+y^2+z^2+2x+3y+4z=0$
• B. $x^2+y^2+z^2-2x+3y+4z=0$
• C. $x^2+y^2+z^2-2x-3y-4z=0$
• D. $x^2+y^2+z^2=27$

Answer 1

The correct option is C $x^2+y^2+z^2-2x-3y-4z=0$

$A=(2,0,0),B=(0,3,0),C=(0,0,4)$
Now equation of sphere passes through origin is given by,
$x^2+y^2+z^2+ax+by+cz=0$
Since it also passes through $A,B,C$
$4+2a=0\Rightarrow a=-2$
and $9+3b=0\Rightarrow b=-3$
$16+4c=0\Rightarrow c=-4$
Hence, equation of the sphere is
$x^2+y^2+z^2-2x-3y-4z=0$
Therefore the number of possible sphere through given points is $1$.