Question

The equation of sphere passing through points $(1,0,0),(0,1,0)$ and $(0,0,1)$ and having minimum radius, is

• A. $3(x^2+y^2+z^2)-x-y-z-1=0$
• B. $3(x^2+y^2+z^2)-2(x+y+z)-1=0$
• C. $3(x^2+y^2+z^2)+2(x+y+z)+1=0$
• D. $3(x^2+y^2+z^2)+x+y+z+1=0$

Answer 1

The correct option is B $3(x^2+y^2+z^2)-2(x+y+z)-1=0$

Let the equation of sphere be
$x^2+y^2+z^2+2ux+2vy+2wz+d=0\cdots \left(i\right)$
As the sphere passes through points $(1,0,0),(0,1,0)$ and $(0,0,1),$ we get
$1+2u+d=0,1+2v+d=0$ and $1+2w+d=0$
$\Rightarrow u=v=w=-\frac{1}{2}(d+1)$
If $R$ is the radius of sphere,
$\Rightarrow R^2=u^2+v^2+w^2-d=\frac{3}{4}(d+1{)}^2-d=\frac{3}{4}[d^2+2d+1-\frac{4}{3}d]=\frac{3}{4}[{(d+\frac{1}{3})}^2+\frac{8}{9}]$
So, for minimum radius, $d+\frac{1}{3}=0$
$\Rightarrow d=-\frac{1}{3}\Rightarrow u=v=w=-\frac{1}{2}(1-\frac{1}{3})=-\frac{1}{3}$
Putting these values in equation $\left(i\right),$ we get
$x^2+y^2+z^2-\frac{2}{3}(x+y+z)-\frac{1}{3}=0$
$\Rightarrow 3(x^2+y^2+z^2)-2(x+y+z)-1=0$