Question

The smallest radius of the sphere passing through $(1,0,0),(0,1,0)$ and $(0,0,1)$ is

• A. $\sqrt{\frac{2}{3}}$
• B. $\sqrt{\frac{3}{8}}$
• C. $\sqrt{\frac{5}{6}}$
• D. $\sqrt{\frac{5}{12}}$

Answer 1

Answer: (A)

$\sqrt{\frac{2}{3}}$

Let the sphere be $x^2+y^2+z^2+2ux+2vy+2wz+d=0$

It passes through $(1,0,0)\Rightarrow 1+2u+d=0$ ....(1)

It passes through $(0,1,0)\Rightarrow 1+2v+d=0$ ....(2)

It passes through $(0,0,1)\Rightarrow 1+2w+d=0$ ...(3)

$\therefore u=v=w=-\frac{1+d}{2}$

Radius of the sphere $=\sqrt{u^2+v^2+w^2-d}$

$=\frac{1}{2}\sqrt{3{(1+d)}^2-4d}=\frac{1}{2}\sqrt{3d^2+2d+3}$

Now, $3d^2+2d+3\ge \frac{4\times 3\times 3-{\left(2\right)}^2}{4\times 3}=\frac{32}{12}=\frac{8}{3}$

$\therefore$ radius $\ge \frac{1}{2}\times \sqrt{\frac{8}{3}}=\sqrt{\frac{2}{3}}$