Show that the plane $2x-2y+z+12=0$ touches the sphere $x^2+y^2+z^2-2x-4y+2z+3=0$. Find the point of contact.

If the plane touches the sphere, then the perpendicular distance from the centre of the sphere to the plane should be equal to the radius of the given sphere.

$x^2+y^2+z^2-2x-4y+2z+3=0 \\=> (x-1)^2 + (y-2)^2 + (z+1)^2 = 9$

Thus centre of sphere is $(1,2,-1$ and radius is $3$.

Now, the distance from $(1,2,-1)$ to the plane is

$\frac{1.2 + 2.(-2) + (-1).1 + 12}{\sqrt{2^2 + (-2)^2 + 1^2}} = \frac{9}{3} = 3$

which is equal to the radius of the sphere. Hence, the plane touches the sphere.

The equation of the line joining centre to the point of intersection can be written as:

$\frac{x-1}{2} = \frac{y-2}{-2} = \frac{z-1}{1} = r$

Any point of this line can be represented as $(2r+1, -2r+2, r-1)$

Putting this in the equation of the plane, we get $r = -1$. Hence, the point of intersection is $(-1, 4, -2)$

answered Nov 11, 2017 by (1,920 points)