A circle cuts the chord on x-axis of length `4a`. If this circle cuts the y-axis at a point whose distance from origin is `2b`. Locus of its centre is (A) Ellipse (B) Parabola (C) Hyperbola (D) Straight line

Options

A. A hyperbola
B. A parabola
C. A straight line
D. An ellipse

Answer

To solve the problem step by step, we will analyze the given information and derive the locus of the center of the circle.

### Step-by-Step Solution:

1. **Understanding the Circle's Properties**:
- The circle cuts a chord on the x-axis with a length of \(4a\). This means that the endpoints of the chord can be represented as \((-2a, 0)\) and \((2a, 0)\).

2. **Finding the Midpoint of the Chord**:
- The midpoint of the chord, which is also the projection of the center of the circle onto the x-axis, is at the origin \((0, 0)\).

3. **Identifying the Center of the Circle**:
- Let the center of the circle be \((h, k)\). The distance from the center to the chord (which lies on the x-axis) is equal to \(k\).

4. **Using the Distance Formula**:
- The radius \(r\) of the circle can be calculated using the distance from the center to one of the endpoints of the chord:
\[ r^2 = (h - 2a)^2 + k^2 \]
- Similarly, the distance from the center to the other endpoint is:
\[ r^2 = (h + 2a)^2 + k^2 \]

5. **Setting Up the Equation**:
- Since both expressions equal \(r^2\), we can set them equal to each other:
\[ (h - 2a)^2 + k^2 = (h + 2a)^2 + k^2 \]
- Simplifying this gives:
\[ (h - 2a)^2 = (h + 2a)^2 \]

6. **Expanding and Simplifying**:
- Expanding both sides:
\[ h^2 - 4ah + 4a^2 = h^2 + 4ah + 4a^2 \]
- Canceling \(h^2\) and \(4a^2\) from both sides, we have:
\[ -4ah = 4ah \]
- This simplifies to:
\[ 8ah = 0 \]
- Thus, \(h = 0\) or \(a = 0\). Since \(a\) cannot be zero (as it represents a length), we conclude \(h = 0\).

7. **Finding the y-coordinate**:
- The circle cuts the y-axis at a point whose distance from the origin is \(2b\). This means the center of the circle has coordinates \((0, k)\) where \(|k| = 2b\).

8. **Locus of the Center**:
- Since \(h = 0\) and \(k\) can take values \(2b\) or \(-2b\), the locus of the center of the circle is a vertical line at \(x = 0\), which is a straight line.

### Final Answer:
The locus of the center of the circle is a straight line.