Question

If any tangent plane to the sphere $(x-a{)}^2+(y-b{)}^2+(z-c{)}^2=r^2$ makes intercepts $a,b,c$ with the coordinate axes at $A,B,C$ and $P$ be the center of the sphere. Then

• A. $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{4}{r^2}$
• B. $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{1}{r^2}$
• C. Volume of the tetrahedron $PABC=\frac{abc}{3}$
• D. Volume of the tetrahedron $PABC=\frac{abc}{6}$

Answer 1

Answer: (A), (C)

The correct options are
A $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{4}{r^2}$
C Volume of the tetrahedron $PABC=\frac{abc}{3}$

Equation of the plane is $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$
Distance of the plane from the center of the sphere is
$\frac{2}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}}=r\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{4}{r^2}$
Now in triangle $ABC$
Distance of $C$ from $AB$
$AB:\frac{x-a}{a}=\frac{y-0}{-b}=\frac{z-0}{0}=k$
any point on the line $\left(a\right(k+1),-bk,0)$
Solving the point with the point $C$
$k=\frac{-a^2}{a^2+b^2}$
So perpendicular distance will be
$=\sqrt{\frac{a^2{b}^2}{a^2+b^2}+c^2}=\sqrt{\frac{a^2{b}^2+a^2{c}^2+b^2{c}^2}{a^2+b^2}}=\frac{2abc}{r\sqrt{(a^2+b^2)}}$
$AB=\sqrt{a^2+b^2}$
$ar.△ABC=\frac{1}{2}\frac{2abc}{r\sqrt{(a^2+b^2)}}\cdot \sqrt{a^2+b^2}=\frac{abc}{r}$
perpendicular distance between plane $ABC$ and $P$
$=r$
$\therefore$ volume $=\frac{r}{3}\frac{abc}{r}=\frac{abc}{3}$