The end A of diameter AB of a circle is (1, 1) and B lies on the line `x+y-3=0`. The locus of the centre of the circle is

Options

A. `x-y=1`
B. `x+y=1`
C. `2x+2y-5=0`
D. `2x-2y-5=0`

Answer

To find the locus of the center of the circle with diameter AB, where point A is given as (1, 1) and point B lies on the line \(x + y - 3 = 0\), we can follow these steps:

### Step 1: Identify the coordinates of points A and B
- Point A is given as \(A(1, 1)\).
- Point B has coordinates \(B(L, M)\) where \(L\) and \(M\) satisfy the line equation \(x + y - 3 = 0\).

### Step 2: Write the line equation in terms of L and M
- From the line equation \(x + y - 3 = 0\), we can express it as:
\[ L + M - 3 = 0 \quad \Rightarrow \quad L + M = 3 \]

### Step 3: Find the coordinates of the center O of the circle
- The center O of the circle, which is the midpoint of diameter AB, can be calculated using the midpoint formula:
\[ O\left(\frac{1 + L}{2}, \frac{1 + M}{2}\right) \]
- Let the coordinates of the center O be \(O(h, k)\), where:
\[ h = \frac{1 + L}{2} \quad \text{and} \quad k = \frac{1 + M}{2} \]

### Step 4: Express L and M in terms of h and k
- From the equation for \(h\):
\[ L = 2h - 1 \]
- From the equation for \(k\):
\[ M = 2k - 1 \]

### Step 5: Substitute L and M into the line equation
- Substitute \(L\) and \(M\) into the line equation \(L + M = 3\):
\[ (2h - 1) + (2k - 1) = 3 \]
- Simplifying this gives:
\[ 2h + 2k - 2 = 3 \quad \Rightarrow \quad 2h + 2k = 5 \]

### Step 6: Rearrange to find the locus equation
- Dividing the entire equation by 2 gives:
\[ h + k = \frac{5}{2} \]
- In terms of \(x\) and \(y\) (where \(h\) corresponds to \(x\) and \(k\) corresponds to \(y\)), the equation becomes:
\[ 2x + 2y - 5 = 0 \]

### Final Answer
- The locus of the center of the circle is given by the equation:
\[ 2x + 2y - 5 = 0 \]