Question

Show that the plane 2x-4y-z+9=0 touches the sphere which passes through (1,1,6) and whose centre is (2,-3,4).Also, find the point of contact

Answer 1

Firstly we should prove, that plane 2x-4y-z+9=0 touches the sphere.

If distance between plane and centre of sphere equals radius of sphere, plane 2x-4y-z+9=0 touches the sphere. So:

$d =\frac{|A·x + B·y + C·z + D|}{\sqrt{A2 + B2 + C2}}$ - formula of distance between plane and point.

$d=\frac{|4+12-4+9|}{\sqrt{21}}=\sqrt{21}$

The radius equals: $\sqrt{(2-1)^2+(-3-1)^2+(4-6)^2}=\sqrt{21}$

It means, that plane touches the sphere.

The equation of sphere: $(x-2)^2+(y+3)^2+(z-4)^2=21$

The equation of tangent plane at the point ($x_0,y_0,z_0$) is:

$F'_x(x_0,y_0,z_0)(x-x_0)+F'_y(x_0,y_0,z_0)(y-y_0)+F'_z(x_0,y_0,z_0)(z-z_0)=0$

The equation of tangent plane of this sphere at the point ($x_0,y_0,z_0$) is:

$2(x_0-2)(x-x_0)+2(y_0+3)(y-y_0)+2(z_0-4)(z-z_0)=0$

$2(x_0-2)x+2(y_0+3)y+2(z_0-4)z-2(x_0-2)x_0-2(y_0+3)y_0-2(z_0-4)z_0=0$

To find this tangent point, we should equate the coefficients of the tangent plane equation to coefficients of given plane.

$2(x_0-2)=2 ; \quad x_0=3$

$2(y_0+3)=-4 ; \quad y_0=-5$

$2(z_0-4)=-1;\quad z_0=3.5$

So our point is (3, -5, 3.5)