Question

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?
A. $(1,4,3)$
B. $(3,4,1)$
C. $( - 1, - 4,2)$
D. $( - 1,4, - 2)$

Answer 1

Firstly, we need to show that the given plane is tangential to the given sphere and since we know tangent touches the sphere it means that the shortest distance between the centre and the plane should be equal to the radius. After proving it is tangential to the sphere, we can find the point of contact by finding the line joining centre and point of contact with the help of the cartesian form of an equation.
Eventually finding the point of intersection as the answer.

Complete step by step solution:
According to the question we need to find the point of contact of the given plane and sphere
Now we know, 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.
${x^2} + {y^2} + {z^2} - 2x - 4y + 2z + 3 = 0$
On adding and subtracting terms and rearranging we get,
$\Rightarrow ({x^2} - 2x + 1) + ({y^2} - 4y + 4) + ({z^2} + 2z + 1) + 3 - 1 - 1 - 4 = 0$
By completing the square method, we get
$\Rightarrow {(x - 1)^2} + {(y - 2)^2} + {(z + 1)^2} = 3$
So, the centre of the sphere is $(1,2, - 1)$ and radius is $\sqrt 3$ .
Since the general equation of a sphere is ${x^2} + {y^2} + {z^2} = {r^2}$ where r is the radius of the sphere.
Now, The Distance Between $(1,2, - 1)$ and the plane $2x - 2y + z + 12 = 0$ is
$\Rightarrow D = \dfrac{{Ax + By + Cz + D}}{{\sqrt {{A^2} + {B^2} + {C^2}} }}$
$Ax + By + Cy + D$ being the equation of the plane with x, y and z being the points through which the distance is to measured, which here is the centre $(1,2, - 1)$ .
$\Rightarrow D = \dfrac{{1 \times 2 + 2 \times - 2 + ( - 1) \times 1 + 12}}{{\sqrt {{2^2} + {{( - 2)}^2} + {1^2}} }}$
On simplification we get,
$\Rightarrow D = \dfrac{{2 - 4 - 1 + 12}}{{\sqrt {4 + 4 + 1} }}$
Hence on solving we get,
$\Rightarrow D = \dfrac{9}{{\sqrt 9 }}$
We know that, $\sqrt {{b^2}} = b$ , Hence
$\Rightarrow D = \dfrac{9}{3}$
On division we get,
$\Rightarrow D = 3$
Therefore, the radius of the sphere and the distance ‘D’ between the centre of the sphere and the plane are equal
Hence, the plane is tangential to the sphere.
Now we know that equation of the line joining the centre and the plane’s point of contact can be written in the cartesian form as
$\dfrac{{x - {x_1}}}{a} = \dfrac{{y - {y_1}}}{b} = \dfrac{{z - {z_1}}}{c} = \lambda$
Where $({x_1},{y_1},{z_1})$ is the centre $(1,2, - 1)$ and a, b and c being the coefficients of x, y and z in the equation of the plane?
$\Rightarrow \dfrac{{x - 1}}{2} = \dfrac{{y - 2}}{{ - 2}} = \dfrac{{z + 1}}{1} = \lambda$
$\Rightarrow$ The general point of the above equation can be written as
$(2\lambda + 1,2 - 2\lambda ,\lambda - 1)$ … (1)
By equating all the equations separately to $\lambda$
Substituting these points in the equation of the plane we get
$\Rightarrow 2(2\lambda + 1) - 2(2 - 2\lambda ) + (\lambda - 1) + 12 = 0$
On solving the bracket, we get,
$\Rightarrow 4\lambda + 2 - 4 + 4\lambda + \lambda - 1 + 12 = 0$
On adding like terms, we get,
$\Rightarrow 9\lambda + 9 = 0$
On subtracting 9 from both sides we get,
$\Rightarrow 9\lambda = - 9$
On dividing the equation by 9 we get,
$\Rightarrow \lambda = \dfrac{{ - 9}}{9}$
Hence, we have,
$\Rightarrow \lambda = - 1$
Now, Substituting the value of $\lambda$ in (1), we get
$\Rightarrow (2( - 1) + 1,2 - 2( - 1),( - 1) - 1)$
On simplifying the bracket, we get,
$\Rightarrow ( - 2 + 1,2 + 2, - 1 - 1)$
On solving we have,
$\Rightarrow ( - 1,4, - 2)$
Hence the point of contact is $( - 1,4, - 2)$

Hence, the final answer is D.

Note:
In these types of questions, once we have got our answer, we can check our answer by putting the final point in the equation of the sphere. If the answer is correct, the point should satisfy the equation of the sphere, if it does not satisfy the equation of the sphere our answer is wrong, we need to check the answer again.