Circles are drawn through the point (3,0) to cut an intercept of length 6 units on the negative direction of the x-axis. The equation of the locus of their centres is

Options

A. y=0
B. y=x
C. x=0
D. y=-x

Explanation

To find the equation of the locus of the centers of circles that pass through the point (3, 0) and cut an intercept of length 6 units on the negative direction of the x-axis, we can follow these steps:

### Step 1: Understand the Circle's Equation
The general equation of a circle can be written as:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
where \((g, f)\) represents the coordinates of the center of the circle.

### Step 2: Substitute the Point (3, 0)
Since the circle passes through the point (3, 0), we can substitute these coordinates into the circle's equation:
\[ (3)^2 + (0)^2 + 2g(3) + 2f(0) + c = 0 \]
This simplifies to:
\[ 9 + 6g + c = 0 \]
Rearranging gives us:
\[ c = -9 - 6g \quad \text{(Equation 1)} \]

### Step 3: Determine the Length of the Intercept
The problem states that the circle cuts an intercept of length 6 units on the negative direction of the x-axis. The length of the intercept on the x-axis can be expressed as:
\[ 2\sqrt{g^2 - c} = 6 \]
Dividing both sides by 2 gives:
\[ \sqrt{g^2 - c} = 3 \]
Squaring both sides results in:
\[ g^2 - c = 9 \quad \text{(Equation 2)} \]

### Step 4: Substitute Equation 1 into Equation 2
Now, we can substitute the expression for \(c\) from Equation 1 into Equation 2:
\[ g^2 - (-9 - 6g) = 9 \]
This simplifies to:
\[ g^2 + 9 + 6g = 9 \]
Subtracting 9 from both sides gives:
\[ g^2 + 6g = 0 \]

### Step 5: Factor the Equation
Factoring out \(g\) from the equation:
\[ g(g + 6) = 0 \]
This gives us two possible solutions:
\[ g = 0 \quad \text{or} \quad g = -6 \]

### Step 6: Find the Locus of the Centers
1. If \(g = 0\), the center of the circle is at \(x = 0\).
2. If \(g = -6\), the center of the circle would be at \(x = -6\).

However, since we are looking for the locus of the centers, we can express this as:
\[ x + 3 = 0 \quad \text{(for } g = -6\text{)} \]
Thus, the locus of the centers of the circles is:
\[ x = 0 \quad \text{(for } g = 0\text{)} \]

### Final Answer
The equation of the locus of their centers is:
\[ x = 0 \]