A circle touches x-axis and cuts off constant length 2p from y-axis then the locus of its centre is

Options

A. `x^(2)+y^(2)=p^(2)`
B. `x^(2)-y^(2)=p^(2)`
C. `y^(2)-x^(2)=p^(2)`
D. `x^(2)+y^(2)=4p^(2)`

Answer

To find the locus of the center of a circle that touches the x-axis and cuts off a constant length of \(2p\) from the y-axis, we will follow these steps:

### Step-by-Step Solution:

1. **Understanding the Circle's Properties**:
- Let the center of the circle be \((-h, -k)\).
- Since the circle touches the x-axis, the radius \(r\) is equal to the absolute value of the y-coordinate of the center, which is \(|k|\).

2. **Equation of the Circle**:
- The general equation of the circle can be expressed as:
\[ (x + h)^2 + (y + k)^2 = k^2 \]
- Expanding this, we get:
\[ x^2 + 2hx + h^2 + y^2 + 2ky + k^2 = k^2 \]
- Simplifying, we have:
\[ x^2 + y^2 + 2hx + 2ky + h^2 = 0 \]

3. **Finding the Y-Intercept**:
- The y-intercept of the circle is given as \(2p\). The formula for the y-intercept in terms of the circle's coefficients is:
\[ 2\sqrt{f^2 - c} \]
- Here, \(f = k\) and \(c = h^2\). Thus, we have:
\[ 2p = 2\sqrt{k^2 - h^2} \]
- Dividing both sides by 2 gives:
\[ p = \sqrt{k^2 - h^2} \]

4. **Squaring Both Sides**:
- Squaring both sides of the equation \(p = \sqrt{k^2 - h^2}\) results in:
\[ p^2 = k^2 - h^2 \]
- Rearranging this gives:
\[ k^2 - h^2 = p^2 \]

5. **Substituting Coordinates**:
- Let \(h = x\) and \(k = y\) (where \((x, y)\) are the coordinates of the center of the circle).
- Substituting these into the equation gives:
\[ y^2 - x^2 = p^2 \]

6. **Final Locus Equation**:
- The equation \(y^2 - x^2 = p^2\) represents the locus of the center of the circle.

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