Question

A plane $x+y+z=1$ intersects with a sphere which passes through the origin. The image of the center of the sphere about the plane lies on the line $6x+8=6y+8=-3z-4$. If the area of the cross section is $16\pi$, then

• A. Radius of the sphere is $\sqrt{\frac{73}{3}}$
• B. Coordinates of the center is $(2+\sqrt{\frac{37}{18}},2+\sqrt{\frac{37}{18}},2-\sqrt{\frac{74}{9}})$
• C. Coordinates of the center is $(2-\sqrt{\frac{37}{18}},2-\sqrt{\frac{37}{18}},2+\sqrt{\frac{74}{9}})$
• D. Radius of the sphere is $2\sqrt{\frac{73}{3}}$

Answer 1

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

Coordinates of the center is $(2-\sqrt{\frac{37}{18}},2-\sqrt{\frac{37}{18}},2+\sqrt{\frac{74}{9}})$


Plane $P:x+y+z=1$
Line $L:6x+8=6y+8=-3z-4$
Line and plane both are parallel to each other.
So, distance between plane and line is $\left|\frac{-\frac{4}{3}-\frac{4}{3}-\frac{4}{3}-1}{\sqrt{3}}\right|=\frac{5\sqrt{3}}{3}$

Let the coordinates of the center of sphere is $A\equiv (a,b,c)$
$\therefore \left|\frac{a+b+c-1}{\sqrt{3}}\right|=\frac{5\sqrt{3}}{3}\Rightarrow a+b+c=6\cdots \left(1\right)$

Equation of the line intersecting $L$ perpendicularly and passing through $(a,b,c)$ is $L_1:x-a=y-b=z-c=k$
Any point on the line $L$ is
$(\frac{\lambda -8}{6},\frac{\lambda -8}{6},-\frac{\lambda +4}{3})$
This point lies on the line $L_1$ also.
$\therefore a=b,\lambda =2(b-c)$

Coordinates of intersecting point of $L$ and $L_1$ will be
$C\equiv (\frac{b-c-4}{3},\frac{b-c-4}{3},\frac{2(c-b)-4}{3})$

Intersecting point of $L_1$ to the plane $P$ is $D(k+a,k+b,k+c)$
$D$ lies on the plane $P$.
So, $k+a+k+b+k+c=1$
$\Rightarrow 3k+a+b+c=1$
$\Rightarrow k=\frac{-5}{3}\cdots \text{(From (1))}$
$D\equiv (a-\frac{5}{3},b-\frac{5}{3},c-\frac{5}{3})$

$D$ is the mid point of $AC$.
Radius of the sphere is $\sqrt{a^2+a^2+c^2}$
So, $\sqrt{2a^2+c^2}=\sqrt{4^2+\frac{25}{3}}$
$\Rightarrow 2a^2+c^2=\frac{73}{3}\cdots \left(2\right)$
From $\left(1\right)\&\left(2\right)$,
$a=b=2\pm \sqrt{\frac{37}{18}},c=2\mp \sqrt{\frac{74}{9}}$