If the radius of the sphere `x^2+y^2+z^2-6x-8y+10z+lambda=0` is unity. What is the value of `lambda` ?

Options

A. 49
B. 7
C. `-49`
D. `-7`

Answer

To find the value of \( \lambda \) such that the radius of the sphere given by the equation

\[ x^2 + y^2 + z^2 - 6x - 8y + 10z + \lambda = 0 \]

is unity (1), we can follow these steps:

### Step 1: Rewrite the equation of the sphere
The general equation of a sphere can be expressed in the form:

\[ x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d = 0 \]

where \( R \) is the radius, and \( (u, v, w) \) are the coordinates of the center of the sphere.

### Step 2: Identify coefficients
From the given equation, we can identify the coefficients:

- \( 2u = -6 \) → \( u = -3 \)
- \( 2v = -8 \) → \( v = -4 \)
- \( 2w = 10 \) → \( w = 5 \)
- \( d = \lambda \)

### Step 3: Use the formula for the radius
The radius \( R \) of the sphere can be calculated using the formula:

\[ R = \sqrt{u^2 + v^2 + w^2 - d} \]

Given that the radius \( R = 1 \), we can substitute the values of \( u, v, w \), and \( d \) into the formula:

\[ 1 = \sqrt{(-3)^2 + (-4)^2 + (5)^2 - \lambda} \]

### Step 4: Calculate the squares
Calculating the squares:

\[ (-3)^2 = 9, \quad (-4)^2 = 16, \quad (5)^2 = 25 \]

So, we have:

\[ 1 = \sqrt{9 + 16 + 25 - \lambda} \]

### Step 5: Simplify the equation
Adding the squares:

\[ 1 = \sqrt{50 - \lambda} \]

### Step 6: Square both sides
Squaring both sides gives:

\[ 1^2 = 50 - \lambda \]

This simplifies to:

\[ 1 = 50 - \lambda \]

### Step 7: Solve for \( \lambda \)
Rearranging the equation to solve for \( \lambda \):

\[ \lambda = 50 - 1 \]

Thus,

\[ \lambda = 49 \]

### Conclusion
The value of \( \lambda \) is \( 49 \).

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