A circle touches the x-axis and also touches the circle with centre (0,3) and radius 2. The locus of the centre of the circle is

Options

A. a circle
B. a parabola
C. an ellipse
D. a hyperbola

Answer

To solve the problem, we need to find the locus of the center of a circle that touches the x-axis and also touches another circle with center (0, 3) and radius 2.

### Step-by-Step Solution:

1. **Understanding the Circle Touching the x-axis**:
- Let the center of the circle we are looking for be \( C(h, k) \) and its radius be \( r \).
- Since the circle touches the x-axis, the distance from the center to the x-axis must equal the radius. Therefore, we have:
\[ k = r \]

2. **Understanding the Circle Touching Another Circle**:
- The second circle has center \( C_2(0, 3) \) and radius \( R_2 = 2 \).
- For the two circles to touch each other, the distance between their centers must equal the sum of their radii:
\[ \text{Distance}(C, C_2) = r + R_2 \]
- This can be expressed mathematically as:
\[ \sqrt{(h - 0)^2 + (k - 3)^2} = r + 2 \]

3. **Substituting the Radius**:
- Since we have \( k = r \), we can substitute \( r \) with \( k \):
\[ \sqrt{h^2 + (k - 3)^2} = k + 2 \]

4. **Squaring Both Sides**:
- Squaring both sides to eliminate the square root gives us:
\[ h^2 + (k - 3)^2 = (k + 2)^2 \]

5. **Expanding Both Sides**:
- Expanding the left side:
\[ h^2 + (k^2 - 6k + 9) = h^2 + k^2 - 6k + 9 \]
- Expanding the right side:
\[ (k^2 + 4k + 4) \]

6. **Setting the Equation**:
- Now we set the expanded forms equal to each other:
\[ h^2 + k^2 - 6k + 9 = k^2 + 4k + 4 \]

7. **Simplifying the Equation**:
- Cancel \( k^2 \) from both sides:
\[ h^2 - 6k + 9 = 4k + 4 \]
- Rearranging gives:
\[ h^2 = 10k - 5 \]

8. **Identifying the Locus**:
- The equation \( h^2 = 10k - 5 \) represents a parabola in the \( hk \)-plane.

### Conclusion:
The locus of the center of the circle is a parabola described by the equation:
\[ h^2 = 10k - 5 \]