What is the centre of the sphere `ax^2+by^2+cz^2-6x=0` if the radius is 1 unit?

Options

A. A) `(0,0,0)`
B. B) `(1,0,0)`
C. C) `(3,0,0)`
D. D) Cannot be determined as values of a,b,c are unknown

Answer

To find the center of the sphere given by the equation \( ax^2 + by^2 + cz^2 - 6x = 0 \) with a radius of 1 unit, we can follow these steps:

### Step 1: Rewrite the equation
The equation of the sphere can be rewritten in a standard form. Since the coefficients of \( x^2 \), \( y^2 \), and \( z^2 \) must be equal for it to represent a sphere, we set \( a = b = c \). Thus, we can rewrite the equation as:
\[ ax^2 + ay^2 + az^2 - 6x = 0 \]

### Step 2: Divide by \( a \)
Next, we divide the entire equation by \( a \) (assuming \( a \neq 0 \)):
\[ x^2 + y^2 + z^2 - \frac{6}{a}x = 0 \]

### Step 3: Rearrange the equation
Rearranging gives us:
\[ x^2 - \frac{6}{a}x + y^2 + z^2 = 0 \]

### Step 4: Complete the square for \( x \)
To complete the square for the \( x \) terms, we take the coefficient of \( x \), which is \( -\frac{6}{a} \), halve it to get \( -\frac{3}{a} \), and square it to get \( \left(\frac{3}{a}\right)^2 \):
\[ x^2 - \frac{6}{a}x + \left(\frac{3}{a}\right)^2 - \left(\frac{3}{a}\right)^2 + y^2 + z^2 = 0 \]
This simplifies to:
\[ \left(x - \frac{3}{a}\right)^2 + y^2 + z^2 = \left(\frac{3}{a}\right)^2 \]

### Step 5: Compare with the standard equation of a sphere
The standard equation of a sphere is given by:
\[ (x - p)^2 + (y - q)^2 + (z - r)^2 = R^2 \]
From our equation, we can identify:
- Center \( (p, q, r) = \left(\frac{3}{a}, 0, 0\right) \)
- Radius \( R = \frac{3}{a} \)

### Step 6: Set the radius equal to 1
Since we know the radius is 1 unit, we set:
\[ \frac{3}{a} = 1 \]
Solving for \( a \):
\[ a = 3 \]

### Step 7: Find the center
Substituting \( a = 3 \) back into the center coordinates:
\[ \text{Center} = \left(\frac{3}{3}, 0, 0\right) = (1, 0, 0) \]

### Final Answer
Thus, the center of the sphere is:
\[ \boxed{(1, 0, 0)} \]