Question

A sphere touches all the coordinate planes at $A,B,C$ and lies in the first octant. Given, the coordinates of $A$ is $(a,a,0)$. Let $P$ be a point on the sphere so that volume of $PABC$ is maximum, then

• A. Equation of the line passing through the center of the sphere and intersecting the sphere at $P$ also, is $x=y=z$
• B. Coordinates of $P$ is $(\frac{a(1+\sqrt{3})}{\sqrt{3}},\frac{a(1+\sqrt{3})}{\sqrt{3}},\frac{a(1+\sqrt{3})}{\sqrt{3}})$
• C. Maximum volume of $PABC$ is $\frac{a^3(1+\sqrt{3})}{6}$
• D. Area of the cross-section of the sphere and the plane $ABC$ is $\frac{2}{3}\pi a^2$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A Equation of the line passing through the center of the sphere and intersecting the sphere at $P$ also, is $x=y=z$
B Coordinates of $P$ is $(\frac{a(1+\sqrt{3})}{\sqrt{3}},\frac{a(1+\sqrt{3})}{\sqrt{3}},\frac{a(1+\sqrt{3})}{\sqrt{3}})$
C Maximum volume of $PABC$ is $\frac{a^3(1+\sqrt{3})}{6}$
D Area of the cross-section of the sphere and the plane $ABC$ is $\frac{2}{3}\pi a^2$

The sphere touches coordinate planes $(x,y,z)$ at $A,B,C$ respectively.
So, the radius of the sphere will be $a$ and center will be $(a,a,a)$

Therefore, equation of the sphere is
$(x-a{)}^2+(y-a{)}^2+(z-a{)}^2=a^2$
So, Plane passing through $A(a,a,0),B(a,0,a),C(0,0,a)$ is $x+y+z=2a$
Distance of the center of sphere from the plane
$=\frac{a+a+a-2a}{\sqrt{3}}=\frac{\sqrt{3}a}{3}$

For maximum volume, the distance of $P$ should be maximum from the plane.
So, the maximum distance of $P$ from the plane $=a+\frac{\sqrt{3}a}{3}=\frac{\sqrt{3}a+3a}{3}$

Line joining the point $P$ to the center of the sphere i.e., $(a,a,a)$ should be perpendicular to plane,
Therefore, equation of the line is $\frac{x-a}{1}=\frac{y-a}{1}=\frac{z-a}{1}$
$\Rightarrow x=y=z$

Now, $AB=BC=CA=a\sqrt{2}$
Distance of $P$ from plane $ABC$ is $h=\frac{\sqrt{3}a+3a}{3}$
So, the volume of the tetrahedron $PABC$ is
$\frac{\text{area of the base}\times \text{height}}{3}=\frac{\sqrt{3}(a\sqrt{2}{)}^2}{4}\times \frac{h}{3}=\frac{a^3(1+\sqrt{3})}{6}$

Coordinates of $P$ is $(\frac{a(1+\sqrt{3})}{\sqrt{3}},\frac{a(1+\sqrt{3})}{\sqrt{3}},\frac{a(1+\sqrt{3})}{\sqrt{3}})$

Area of the cross section:
Radius of the cross section is$\cos {30}^{\circ }=\frac{a}{\sqrt{2}r}$
$r=\sqrt{\frac{2}{3}}a\therefore \text{area}=\frac{2\pi a^2}{3}$