The locus of the centre of the circle which makes equal intercepts on the lines `x+y=1 and x+y=5` is

Options

A. `x-y=2`
B. `x+y=6`
C. `x+y=3`
D. `x-y=0`

Answer

To find the locus of the center of the circle that makes equal intercepts on the lines \(x + y = 1\) and \(x + y = 5\), we can follow these steps:

### Step 1: Understand the problem
We need to find the locus of the center of a circle that makes equal intercepts on the two given lines. The lines are \(L_1: x + y = 1\) and \(L_2: x + y = 5\).

### Step 2: Define the center of the circle
Let the center of the circle be denoted as \(C(h, k)\).

### Step 3: Calculate the distance from the center to the lines
The distance \(D_1\) from the point \(C(h, k)\) to the line \(L_1: x + y - 1 = 0\) is given by the formula:
\[ D_1 = \frac{|h + k - 1|}{\sqrt{1^2 + 1^2}} = \frac{|h + k - 1|}{\sqrt{2}} \]

Similarly, the distance \(D_2\) from the point \(C(h, k)\) to the line \(L_2: x + y - 5 = 0\) is:
\[ D_2 = \frac{|h + k - 5|}{\sqrt{1^2 + 1^2}} = \frac{|h + k - 5|}{\sqrt{2}} \]

### Step 4: Set the distances equal
Since the circle makes equal intercepts on both lines, we have:
\[ D_1 = D_2 \]
This leads to:
\[ \frac{|h + k - 1|}{\sqrt{2}} = \frac{|h + k - 5|}{\sqrt{2}} \]
We can simplify this to:
\[ |h + k - 1| = |h + k - 5| \]

### Step 5: Solve the absolute value equation
This absolute value equation can be split into two cases:

**Case 1:**
\[ h + k - 1 = h + k - 5 \]
This simplifies to:
\[ 1 = 5 \quad \text{(not possible)} \]

**Case 2:**
\[ h + k - 1 = -(h + k - 5) \]
This simplifies to:
\[ h + k - 1 = -h - k + 5 \]
Bringing all terms involving \(h\) and \(k\) to one side gives:
\[ h + k + h + k = 5 + 1 \]
\[ 2h + 2k = 6 \]
Dividing by 2:
\[ h + k = 3 \]

### Step 6: Write the locus equation
The locus of the center of the circle is given by:
\[ x + y = 3 \]
where we have substituted \(h\) with \(x\) and \(k\) with \(y\).

### Final Answer
The locus of the center of the circle is:
\[ \boxed{x + y = 3} \]