Question

The equation of the sphere inscribed in a tetrahedron, whose faces are $x=0,y=0,z=0$ and $x+2y+2z=1$ is

• A. $32(x^2+y^2+z^2)+8(x+y+z)+1=0$
• B. $32(x^2+y^2+z^2)-8(x+y+z)-1=0$
• C. $32(x^2+y^2+z^2)-8(x+y+z)+1=0$
• D. None of these

Answer 1

The correct option is C $32(x^2+y^2+z^2)-8(x+y+z)+1=0$

Let $(a,b,c)$ be the Centre and $r,$ the radius of the sphere.

The sphere is inscribed in the tetrahedron, hence the length of the perpendicular from the centre $(a,b,c)$ upon each of the faces $=$ radius of the sphere

$\therefore \frac{a}{1}=\frac{b}{1}=\frac{c}{1}=\frac{1-a-2b-2c}{\sqrt{1+4+4}}=r$

i.e., $a=b=c=\frac{1-a-2b-2c}{3}=r$ ...(1)

$\therefore$ From (1), we get

$3a=1-a-2b-2c$ ...(2)

and, $a=b=c$

$\therefore 3a=1-a-2a-2a\Rightarrow 8a=1$

$\therefore a=\frac{1}{8}$

and, then $r=a=\frac{1}{8}$

$\therefore$ Centre is $(\frac{1}{8},\frac{1}{8},\frac{1}{8})$ and radius $=\frac{1}{8}$

Hence, the required sphere is

${(x-\frac{1}{8})}^2+{(y-\frac{1}{8})}^2+{(z-\frac{1}{8})}^2={\left(\frac{1}{8}\right)}^2$

$\Rightarrow 32(x^2+y^2+z^2)-8(x+y+z)+1=0$