A circle is the locus of a point in a plane such that its distance from a fixed point in the plane is constant. Anologously, a sphere is the locus of a point in space such that its distance from a fixed point in space in constant. The fixed point is called the centre and the constant distance is called the radius of the circle/sphere. In anology with the equation of the circle `|z-c|=a`, the equation of a sphere of radius a is `|r-c|=a`, where c is the position vector of the centre and r is the position vector of any point on the surface of the sphere. In Cartesian system, the equation of the sphere, with centre at `(-g, -f, -h)` is `x^2+y^2+z^2+2gx+2fy+2hz+c=0` and its radius is `sqrt(f^2+g^2+h^2-c)`. Q. Equation of the sphere having centre at `(3, 6, -4)` and touching the plane `rcdot(2hat(i)-2hat(j)-hat(k))=10` is `(x-3)^2+(y-6)^2+(z+4)^2=k^2`, where k is equal to

Options

A. `3`
B. `4`
C. `6`
D. `sqrt(17)`

Answer

To solve the problem, we need to find the radius \( k \) of the sphere that has its center at the point \( (3, 6, -4) \) and touches the plane given by the equation \( 2x - 2y - z = 10 \).

### Step-by-Step Solution:

1. **Identify the Center of the Sphere**:
The center of the sphere is given as \( (3, 6, -4) \).

2. **Identify the Plane Equation**:
The equation of the plane is \( 2x - 2y - z = 10 \). We can rewrite this in the standard form \( Ax + By + Cz + D = 0 \) as:
\[ 2x - 2y - z - 10 = 0 \]
Here, \( A = 2 \), \( B = -2 \), \( C = -1 \), and \( D = -10 \).

3. **Find the Normal Vector of the Plane**:
The normal vector \( \vec{n} \) of the plane is given by the coefficients of \( x, y, z \) in the plane equation:
\[ \vec{n} = (2, -2, -1) \]

4. **Calculate the Perpendicular Distance from the Center to the Plane**:
The distance \( d \) from a point \( (x_0, y_0, z_0) \) to the plane \( Ax + By + Cz + D = 0 \) is given by the formula:
\[ d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \]
Substituting the center \( (3, 6, -4) \) into the formula:
\[ d = \frac{|2(3) - 2(6) - (-4) - 10|}{\sqrt{2^2 + (-2)^2 + (-1)^2}} \]
Simplifying the numerator:
\[ = \frac{|6 - 12 + 4 - 10|}{\sqrt{4 + 4 + 1}} = \frac{|-12|}{\sqrt{9}} = \frac{12}{3} = 4 \]

5. **Determine the Radius \( k \)**:
Since the sphere touches the plane, the radius \( k \) of the sphere is equal to the distance \( d \) calculated above. Therefore:
\[ k = 4 \]

6. **Final Equation of the Sphere**:
The equation of the sphere can be expressed as:
\[ (x - 3)^2 + (y - 6)^2 + (z + 4)^2 = k^2 \]
where \( k = 4 \), so \( k^2 = 16 \).

### Conclusion:
The value of \( k \) is \( 4 \).