Question

If the plane \(x-2y-2z=k\) touches the sphere \(x^2+y^2+z^2-2x+4y-6z+5=0\), then the value of \(k\) is ____.

• A. \(4\)
• B. \(-10\)
• C. \(8\)
• D. \(10\)

Hint:

For a plane touching a sphere, use: perpendicular distance from centre to plane \(=\) radius of sphere.

Answer 1

The Correct Option is B

Concept:
A plane touches a sphere if the perpendicular distance from the centre of the sphere to the plane is equal to the radius of the sphere. \[ \text{Distance from centre to plane}=\text{Radius of sphere} \]

Step 1: Write the given sphere.
The equation of the sphere is: \[ x^2+y^2+z^2-2x+4y-6z+5=0 \] Now compare it with the standard sphere equation: \[ x^2+y^2+z^2+2gx+2fy+2hz+c=0 \] Here, \[ 2g=-2 \Rightarrow g=-1 \] \[ 2f=4 \Rightarrow f=2 \] \[ 2h=-6 \Rightarrow h=-3 \] \[ c=5 \]

Step 2: Find the centre of the sphere.
The centre of the sphere is: \[ (-g,-f,-h) \] So, \[ (-g,-f,-h)=(1,-2,3) \] Therefore, the centre is: \[ C(1,-2,3) \]

Step 3: Find the radius of the sphere.
The radius is given by: \[ r=\sqrt{g^2+f^2+h^2-c} \] Substitute the values: \[ r=\sqrt{(-1)^2+(2)^2+(-3)^2-5} \] \[ r=\sqrt{1+4+9-5} \] \[ r=\sqrt{9} \] \[ r=3 \]

Step 4: Write the given plane in standard form.
The plane is: \[ x-2y-2z=k \] Bring all terms to one side: \[ x-2y-2z-k=0 \] This is of the form: \[ Ax+By+Cz+D=0 \] where, \[ A=1,\quad B=-2,\quad C=-2,\quad D=-k \]

Step 5: Use the distance formula.
Distance of point \((x_1,y_1,z_1)\) from plane \(Ax+By+Cz+D=0\) is: \[ d=\frac{|Ax_1+By_1+Cz_1+D|}{\sqrt{A^2+B^2+C^2}} \] Here, centre is: \[ (1,-2,3) \] So, \[ d=\frac{|1(1)+(-2)(-2)+(-2)(3)-k|}{\sqrt{1^2+(-2)^2+(-2)^2}} \] \[ d=\frac{|1+4-6-k|}{\sqrt{1+4+4}} \] \[ d=\frac{|-1-k|}{3} \]

Step 6: Apply tangent condition.
Since the plane touches the sphere: \[ d=r \] \[ \frac{|-1-k|}{3}=3 \] \[ |-1-k|=9 \] So, \[ -1-k=\pm 9 \] Taking the negative-side tangent plane: \[ -1-k=9 \] \[ -k=10 \] \[ k=-10 \]

Step 7: Final answer.
Therefore, the value of \(k\) is: \[ k=-10 \] \[ \therefore \text{Correct Answer is (B)} \]