The locus of the centre of the circle cutting the circles `x^2+y^2–2x-6y+1=0, x^2 + y^2 - 4x - 10y + 5 = 0` orthogonally is

Options

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

Answer

To find the locus of the center of the circle that cuts the given circles orthogonally, we will follow these steps:

### Step 1: Write the equations of the given circles
The equations of the circles are:
1. \( C_1: x^2 + y^2 - 2x - 6y + 1 = 0 \)
2. \( C_2: x^2 + y^2 - 4x - 10y + 5 = 0 \)

### Step 2: Identify the general form of the circle
The general form of a circle is given by:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
From the equations of the circles, we can identify:
- For \( C_1 \): \( g_1 = -1 \), \( f_1 = -3 \), \( c_1 = 1 \)
- For \( C_2 \): \( g_2 = -2 \), \( f_2 = -5 \), \( c_2 = 5 \)

### Step 3: Use the orthogonality condition
Two circles are said to cut orthogonally if the following condition holds:
\[ g_1g_2 + f_1f_2 = c_1 + c_2 \]
Let the center of the required circle be \( (h, k) \). The equation of the circle can be written as:
\[ x^2 + y^2 + 2hx + 2ky + r^2 = 0 \]
Here, \( g = h \), \( f = k \), and \( c = r^2 \).

### Step 4: Set up the equations for orthogonality
Using the orthogonality condition for both circles:
1. For \( C_1 \):
\[ (-1)(h) + (-3)(k) = 1 + r^2 \quad \text{(1)} \]
2. For \( C_2 \):
\[ (-2)(h) + (-5)(k) = 5 + r^2 \quad \text{(2)} \]

### Step 5: Simplify the equations
From equation (1):
\[ -h - 3k = 1 + r^2 \implies r^2 = -h - 3k - 1 \quad \text{(3)} \]
From equation (2):
\[ -2h - 5k = 5 + r^2 \implies r^2 = -2h - 5k - 5 \quad \text{(4)} \]

### Step 6: Set equations (3) and (4) equal to each other
\[ -h - 3k - 1 = -2h - 5k - 5 \]
Rearranging gives:
\[ h + 2k = 4 \quad \text{(5)} \]

### Step 7: Write the locus equation
The equation \( h + 2k = 4 \) can be rewritten as:
\[ h + 2k - 4 = 0 \]
This represents the locus of the center of the circle that cuts the given circles orthogonally.

### Step 8: Final answer
The locus of the center of the circle is:
\[ x + 2y - 4 = 0 \]