Under what condition does the equations `x^2+y^2+z^2+2ux+2uy+2w+d=0` represent a real sphere?

Options

A. `u^2+v^2+w^2=d^2`
B. `u^2+v^2+w^2ltd`
C. `u^2+v^2+w^2gtd`
D. `u^2+v^2+w^2gtd^2`

Explanation

To determine the condition under which the equation

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

represents a real sphere, we can follow these steps:

### Step 1: Identify the standard form of the sphere equation
The general equation of a sphere in 3D space can be expressed in the form:

\[ (x - h)^2 + (y - k)^2 + (z - l)^2 = r^2 \]

where \((h, k, l)\) is the center of the sphere and \(r\) is the radius.

### Step 2: Rewrite the given equation
We can rearrange the given equation to match the standard form of a sphere. The equation can be rewritten as:

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

### Step 3: Complete the square
To convert the equation into the standard form, we complete the square for each variable:

1. For \(x\):
\[ x^2 + 2ux = (x + u)^2 - u^2 \]

2. For \(y\):
\[ y^2 + 2vy = (y + v)^2 - v^2 \]

3. For \(z\):
\[ z^2 + 2wz = (z + w)^2 - w^2 \]

### Step 4: Substitute back into the equation
Substituting these completed squares back into the equation gives us:

\[ (x + u)^2 - u^2 + (y + v)^2 - v^2 + (z + w)^2 - w^2 + d = 0 \]

This simplifies to:

\[ (x + u)^2 + (y + v)^2 + (z + w)^2 = u^2 + v^2 + w^2 - d \]

### Step 5: Identify the radius condition
For this equation to represent a real sphere, the right-hand side must be positive, as it represents \(r^2\) (the square of the radius). Therefore, we need:

\[ u^2 + v^2 + w^2 - d > 0 \]

### Step 6: Rearranging the inequality
Rearranging this inequality gives us the condition:

\[ u^2 + v^2 + w^2 > d \]

### Conclusion
Thus, the condition under which the given equation represents a real sphere is:

\[ u^2 + v^2 + w^2 > d \]