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: Identify the lines
The given lines are:
1. \(x + y = 1\) (let's call this Line 1)
2. \(x + y = 5\) (let's call this Line 2)

These two lines are parallel since they have the same coefficients for \(x\) and \(y\).

### Step 2: Determine the distance from a point to a line
The formula for the perpendicular distance \(D\) from a point \((h, k)\) to a line \(Ax + By + C = 0\) is given by:
\[ D = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
For our lines, we can rewrite them in the standard form:
- Line 1: \(x + y - 1 = 0\) (here, \(A = 1\), \(B = 1\), \(C = -1\))
- Line 2: \(x + y - 5 = 0\) (here, \(A = 1\), \(B = 1\), \(C = -5\))

### Step 3: Calculate the distances
The distance from the center of the circle \((h, k)\) to Line 1 is:
\[ D_1 = \frac{|h + k - 1|}{\sqrt{1^2 + 1^2}} = \frac{|h + k - 1|}{\sqrt{2}} \]

The distance from the center of the circle \((h, k)\) to Line 2 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 set the distances equal:
\[ \frac{|h + k - 1|}{\sqrt{2}} = \frac{|h + k - 5|}{\sqrt{2}} \]
This simplifies to:
\[ |h + k - 1| = |h + k - 5| \]

### Step 5: Solve the absolute value equation
This gives us two cases to consider:

**Case 1:**
\[ h + k - 1 = h + k - 5 \]
This leads to a contradiction (0 = -4), so this case is not valid.

**Case 2:**
\[ h + k - 1 = -(h + k - 5) \]
This simplifies to:
\[ h + k - 1 = -h - k + 5 \]
Combining like terms:
\[ 2(h + k) = 6 \]
Thus:
\[ h + k = 3 \]

### Step 6: Write the locus
The locus of the center of the circle is given by the equation:
\[ h + k = 3 \]
This can also be expressed as:
\[ x + y = 3 \]

### Final Answer
The locus of the center of the circle which makes equal intercepts on the lines \(x + y = 1\) and \(x + y = 5\) is:
\[ \boxed{x + y = 3} \]