Question

The equation of the sphere circumscribing the

tetrahedron whose faces are $\frac{y}{b}+\frac{z}{c}=0,\frac{z}{c}+\frac{x}{a}=0,\frac{x}{a}+\frac{y}{b}=0$ and $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$ is

• A. $x^2+y^2+z^2-ax-by-cz=0$
• B. $x^2+y^2+z^2-(a^2+b^2+c^2)(\frac{x}{a}+\frac{y}{b}+\frac{z}{c})=0$
• C. $x^2+y^2+z^2=a^2+b^2+c^2$
• D. $x^2+y^2+z^2-\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)=0$

Answer 1

The correct option is A $x^2+y^2+z^2-ax-by-cz=0$

Solving given planes we get the vertices of tetrahedron $(0,0,0),(a,0,0),(0,b,0),(0,0,c)$
Now equation of sphere passing through origin is given by,
$x^2+y^2+z^2+ux+vy+wz=0$
Now this sphere is also passing through other vertices of the tetrahedron,
$\therefore a^2+ua=0\Rightarrow u=-a$

Similarly $v=-b,w=-c$
Hence, required sphere is $x^2+y^2+z^2-ax-by-cz=0$