Question

Find 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 $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$

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

Answer 1

The correct option is A $x^2+y^2+z^2-(a^2+b^2+c^2)(\frac{x}{a}+\frac{y}{b}+\frac{z}{c})=0$

Solution:

$\frac{y}{b}+\frac{z}{c}=\mathrm{0..........}\left(1\right)$

$\frac{z}{c}+\frac{x}{a}=\mathrm{0..........}\left(2\right)$

$\frac{x}{a}+\frac{y}{b}=\mathrm{0..........}\left(3\right)$

$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=\mathrm{1..........}\left(4\right)$

Four planes which are faces of tetrahedron are,

From (1), (2), and (3)

$y=b,z=c$

$\frac{z}{c}+\frac{x}{a}=0$

$\Rightarrow \frac{c}{c}+\frac{x}{a}=0$

$\Rightarrow x=-a$

$(x,y,z)=(-a,b,c)...........\left(5\right)$

From (2), (3), and (4)

$x=a,z=c$

$\frac{x}{a}+\frac{y}{b}=0$

$\Rightarrow \frac{a}{a}+\frac{y}{b}=0$

$\Rightarrow y=-b$

$(x,y,z)=(a,-b,c)...........\left(6\right)$

From (3), (4), and (1)

$x=a,y=b$

$\frac{y}{b}+\frac{z}{c}=0$

$\Rightarrow \frac{b}{b}+\frac{y}{b}=0$

$\Rightarrow z=-c$

$(x,y,z)=(a,b,-c)...........\left(7\right)$

Now,

let the equation of sphere be,

$x^2+y^2+z^2+ux+vy+wz=0$

putting the values from (5), (6), and (7) we get,

$a^2+b^2+c^2-ua+vb+wc=\mathrm{0..............}\left(8\right)$

$a^2+b^2+c^2+ua-vb+wc=\mathrm{0..............}\left(9\right)$

$a^2+b^2+c^2+ua+vb-wc=\mathrm{0..............}\left(10\right)$

Solving equation (8), (9) , (10) we get,

$u=\frac{-a^2-b^2-c^2}{a},v=\frac{-a^2-b^2-c^2}{b},w=\frac{-a^2-b^2-c^2}{c}$

Putting these values in equation of sphere,

$\Rightarrow x^2+y^2+z^2+\left(\frac{-a^2-b^2-c^2}{a}\right)x+\left(\frac{-a^2-b^2-c^2}{b}\right)y+\left(\frac{-a^2-b^2-c^2}{c}\right)z=0$

$\Rightarrow x^2+y^2+z^2+(-a^2-b^2-c^2)(\frac{x}{a}+\frac{y}{b}+\frac{z}{c})=0$

$\Rightarrow x^2+y^2+z^2-(a^2+b^2+c^2)(\frac{x}{a}+\frac{y}{b}+\frac{z}{c})=0$