Question

A sphere $S$ passes through the origin and the image of its centre in the plane $x+y+z=3$ is $(0,0,0).$ Radius of the sphere is

• A. $3\sqrt{2}$
• B. $2\sqrt{3}$
• C. $2\sqrt{2}$
• D. $4$

Answer 1

The correct option is B $2\sqrt{3}$

Let the centre 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 above line is $(k,k,k)$
Midpoint of line joining the centre and origin lies on the plane $x+y+z=3$
$\Rightarrow 3k=3$
$\Rightarrow k=1$
Midpoint $(\frac{x_1}{2},\frac{y_1}{2},\frac{z_1}{2})=(1,1,1)$
$\Rightarrow (x_1,y_1,z_1)=(2,2,2)$
Hence, centre of the sphere is $(2,2,2)$
Radius $=\sqrt{(2-0{)}^2+(2-0{)}^2+(2-0{)}^2}=2\sqrt{3}$