Question

A sphere $S$ which passes through origin and the image of it's center in the plane $x+y+z=3$ is $(0,0,0)$. If $a$ be the area of the cross section made by the plane then

• A. $a=9\pi$
• B. $a=4\pi$
• C. $S:\frac{(x^2+y^2+z^2)}{4}=(x+y+z)$
• D. $S:\frac{(x^2+y^2+z^2)}{2}=(x+y+z)$

Answer 1

Answer: (A), (C)

$S:\frac{(x^2+y^2+z^2)}{4}=(x+y+z)$

Let the center of the sphere be $(x_1,y_1,z_1)$
Its image in the plane is $(0,0,0)$
Line perpendicular to the plane and passing through the centre and origin is,
$\frac{x-0}{1}=\frac{y-0}{1}=\frac{z-0}{1}=k$
Any point on the line,
$(k,k,k)$
Midpoint of line joining the centre and origin lies on the plane,
$x+y+z=3\Rightarrow 3k=3\Rightarrow k=1$
Midpoint
$(\frac{x_1}{2},\frac{y_1}{2},\frac{z_1}{2})=(1,1,1)(x_1,y_1,z_1)=(2,2,2)$
Center of the sphere is $(2,2,2)$
Radius $=\sqrt{2^2+2^2+2^2}=2\sqrt{3}$

Equation of the sphere will be
$S:(x-2{)}^2+(y-2{)}^2+(z-2{)}^2=12x^2+y^2+z^2=4(x+y+z)\frac{(x^2+y^2+z^2)}{4}=(x+y+z)$

Cross section of the intersection will be circle
Distance between centre and midpoint is,
$\sqrt{(2-1{)}^2+(2-1{)}^2+(2-1{)}^2}=\sqrt{3}$

Therefore the radius of the circular cross section,
$=\sqrt{(2\sqrt{3}{)}^2-(\sqrt{3}{)}^2}=3$
Hence, the area is
$a=\pi r^2=9\pi$