A circle touches a given straight line and cuts off a constant length 2d from another straight line perpendicular to the first straight line. The locus of the centre of the circle, is

Options

A. `y^(2)-x^(2)=d^(2)`
B. `x^(2)+y^(2)=d^(2)`
C. `xy=d^(2)`
D. none of these

Explanation

To find the locus of the center of a circle that touches a given straight line and cuts off a constant length from another straight line perpendicular to the first line, we can follow these steps:

### Step 1: Understand the Problem
Let’s denote the given straight line as the x-axis and the other line perpendicular to it as the y-axis. The circle touches the x-axis, which means the distance from the center of the circle to the x-axis is equal to the radius of the circle.

### Step 2: Define the Circle
Let the center of the circle be at the point \((-g, -f)\) in the Cartesian coordinate system. The equation of the circle can be expressed as:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]

### Step 3: Condition for Tangency
Since the circle touches the x-axis, the distance from the center to the x-axis is equal to the radius. The radius can be expressed as \(r = -f\) (since the center is at \(-f\) below the x-axis). The condition for tangency gives us:
\[ c = -g^2 \]

### Step 4: Circle Cuts Off Length on the Y-axis
The circle also cuts off a constant length \(2d\) on the y-axis. The length of the intercept on the y-axis can be calculated using the formula for the intercept:
\[ 2d = 2\sqrt{f^2 - c} \]
Substituting \(c = -g^2\) into the equation:
\[ 2d = 2\sqrt{f^2 + g^2} \]
Thus, we have:
\[ d = \sqrt{f^2 + g^2} \]

### Step 5: Relate \(d\) to \(g\) and \(f\)
From the previous step, we can square both sides to eliminate the square root:
\[ d^2 = f^2 + g^2 \]

### Step 6: Substitute for \(g\) and \(f\)
Since the center of the circle is at \((-g, -f)\), we can replace \(g\) with \(-x\) and \(f\) with \(-y\):
\[ d^2 = (-y)^2 + (-x)^2 \]
This simplifies to:
\[ d^2 = y^2 + x^2 \]

### Step 7: Rearranging the Equation
Rearranging gives us the equation:
\[ y^2 - x^2 = d^2 \]

### Conclusion
Thus, the locus of the center of the circle is given by the equation:
\[ y^2 - x^2 = d^2 \]