Question

Find the equation of the sphere which passes through the points $(1,0,0),(0,1,0)$ and $(0,0,1)$ and whose centre lies on the plane $3x-y+z=2$.

Answer 1

Let the equation of sphere $x^2+y^2+z^2+2ux+2vy+2wz+d=0.....\left(1\right)$
Sphere passes through the points $(1,0,0),(0,1,0)$ and $(0,0,1)$
$\therefore 1^2+0^2+0^2+2u.1+2v.0+2w.0+d=0$
$\therefore 2u+d=1=0.....\left(2\right)$
$\therefore 2v+d+1=0.....\left(3\right)$
and $2w+d+1=0.....\left(4\right)$
$\because$ Centre of sphere $(-u,-v,-w)$
lies on plane $3x-y+z=3$
$\therefore -3u+v-w=3....\left(5\right)$
solve the equation $\left(2\right),\left(3\right),\left(4\right)$ and $\left(5\right)$
We get $u=-2/3,v=-2/3,w=-2/3,d=1/3$
putting these values in equation $\left(1\right)$ we get the equation of sphere
$3(x^2+y^2+z^2)-4(x+y+z)+1=0$