A circle passes through A(1,1) and touches x-axis then the locus of the other end of the diameter through A is

Options

A. `(x+1)^(2)=4y`
B. `(y-1)^(2)=4x`
C. `(x-1)^(2)=4y`
D. `(y+1)^(2)=4x`

Answer

To solve the problem, we need to find the locus of the other end of the diameter of a circle that passes through the point A(1,1) and touches the x-axis.

### Step-by-Step Solution:

1. **Understanding the Circle's Properties**:
- 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.

2. **Let the Center of the Circle be C(h, k)**:
- Since the circle touches the x-axis, the y-coordinate of the center (k) must be equal to the radius (r).
- Therefore, we have \( r = k \).

3. **Finding the Radius**:
- The radius can also be expressed as half the distance from point A(1, 1) to point B(h, k), which is the other end of the diameter.
- The distance AB can be calculated using the distance formula:
\[ AB = \sqrt{(h - 1)^2 + (k - 1)^2} \]
- Therefore, the radius \( r \) is:
\[ r = \frac{1}{2} \sqrt{(h - 1)^2 + (k - 1)^2} \]

4. **Setting Up the Equation**:
- Since the radius \( r \) is equal to \( k \), we can set up the equation:
\[ k = \frac{1}{2} \sqrt{(h - 1)^2 + (k - 1)^2} \]

5. **Squaring Both Sides**:
- To eliminate the square root, we square both sides:
\[ k^2 = \frac{1}{4} \left((h - 1)^2 + (k - 1)^2\right) \]

6. **Expanding the Equation**:
- Expanding the right side:
\[ k^2 = \frac{1}{4} \left((h - 1)^2 + (k^2 - 2k + 1)\right) \]
- This simplifies to:
\[ k^2 = \frac{1}{4}(h^2 - 2h + 1 + k^2 - 2k + 1) \]

7. **Rearranging the Equation**:
- Multiply through by 4 to eliminate the fraction:
\[ 4k^2 = h^2 - 2h + 2 + k^2 - 2k \]
- Rearranging gives:
\[ 3k^2 + 2k - h^2 + 2h - 2 = 0 \]

8. **Finding the Locus**:
- To express this in terms of \( h \) and \( k \), we can isolate \( k \):
\[ k = \frac{1}{3} (h^2 - 2h + 2) \]
- This represents a parabola in the \( hk \)-plane.

9. **Final Equation**:
- From the earlier steps, we can also express it as:
\[ (h - 1)^2 = 4k \]
- Thus, the locus of the other end of the diameter through A(1, 1) is given by:
\[ (x - 1)^2 = 4y \]

### Conclusion:
The locus of the other end of the diameter through A(1,1) is the parabola given by the equation:
\[ (x - 1)^2 = 4y \]