Question

Obtain the equation of the sphere having the circle $x^2+y^2+z^2+10y-4z=8,$ $x+y+z=3$ as a great circle.

• A. $x^2+y^2+z^2-4x+6y-8z+4=0$
• B. $x^2+y^2+z^2-4x+6y-8z+9=0$
• C. $x^2+y^2+z^2-4x-6y-8z+4=0$
• D. $x^2+y^2+z^2-4x-6y-8z+9=0$

Answer 1

The correct option is A $x^2+y^2+z^2-4x+6y-8z+4=0$

Any sphere through the intersection of the given circle

$x^2+y^2+z^2+10y-4z-8=0$ ...(1)

$x+y+z=3$ ...(2)

is $x^2+y^2+z^2+10y-4z-8+k(x+y+z-3)=0$

$\Rightarrow x^2+y^2+z^2+kx+(10+k)y-(4-k)z-8-3k=0$ ...(3)

Now if the circle of intersection of (1) and (2) is a greater circle of sphere (3), then

$C(-\frac{k}{2},-\frac{(10+k)}{2},-\frac{4-k}{2})$ of sphere (3) must lie on the plane (2) i.e., we must have

$-\frac{k}{2}-\left(\frac{10+k}{2}\right)+\frac{4-k}{2}-3=0$

$\Rightarrow -k-10-k+4-k-6=0\Rightarrow -3k-12=0$

$\Rightarrow k=-4$

Substitute $k=-4$ in (3), the required sphere is $x^2+y^2+z^2-4x+6y-8z+4=0$.