Question

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

Answer 1

We are firstly going to find the distance between the given plane and the sphere, the distance says it all, that is, If the distance is equal to the radius that means that the plane touches the sphere. And after that the point of contact can also be found.

Formula used:
The distance between the two points $\left( a,b,c \right)and\left( d,e,f \right)$ is
$s=\sqrt{{{\left( a-d \right)}^{2}}+{{\left( b-e \right)}^{2}}+{{\left( c-f \right)}^{2}}}$
Distance between a plane and a point is given by:
The plane $px+qy+rz+m=0$and point $\left( a,b,c \right)$
$s=\dfrac{\left| \left( p\centerdot a+q\centerdot b+r\centerdot c+m \right) \right|}{\sqrt{{{\left( p \right)}^{2}}+{{\left( q \right)}^{2}}+{{\left( r \right)}^{2}}}}$
For point of contact:
$\vec{u}=\left( a,b,c \right)-\left( p,q,r \right)$

Complete step by step answer:
It is given that the sphere passes through the point $\left( 1,1,6 \right)$ and the center of the sphere is at the point $\left( 2,-3,4 \right)$. The distance of this point from the sphere, is equal to the radius of the sphere, i.e. the distance between the center and the point.If radius of sphere is $r$.

\Rightarrow r=\sqrt{1+16+4} \\\ \Rightarrow r=\sqrt{21} \\\ $$ Now the distance between the given plane $$2x-4y-z+9=0$$ and the center of the sphere is $$\dfrac{\left| \left( 2\centerdot 2+\left( -4 \right)\centerdot \left( -3 \right)+\left( -1 \right)\centerdot 4+9 \right) \right|}{\sqrt{{{\left( 2 \right)}^{2}}+{{\left( -4 \right)}^{2}}+{{\left( -1 \right)}^{2}}}} \\\ \Rightarrow\dfrac{\left| \left( 4+12+-4+9 \right) \right|}{\sqrt{21}} \\\ \Rightarrow\dfrac{21}{\sqrt{21}}=\sqrt{21} \\\ $$ Here, we can see that the distance of the plane and the sphere is exactly equal to the radius of the sphere. This is possible only if the plane passes through the sphere, therefore, we can say that the plane $$2x-4y-z+9=0$$ passes through the sphere with center $$\left( 2,-3,4 \right)$$ and radius$$\sqrt{21}$$. The point of contact is $$\vec{u}=\left( 2,-3,4 \right)-\left( 2,-4,-1 \right) \\\ \therefore \vec{u}=\left( 0,1,5 \right) $$ **Hence, the point of contact is $\left( 0,1,5 \right)$.** **Note:** If a sphere passes through a point, then the radius is equal to its distance from the center. The coefficients of $$x,y,z$$ in the equation for the plane serve as the coordinates and used like that only while solving the question.