Find the radical centre of the following circles
`x^2+y^2-4x-6y+5=0`
`x^2+y^2-2x-4y-1=0`
`x^2+y^2-6x-2y=0`

Answer

To find the radical center of the three given circles, we will follow these steps:

### Step 1: Write the equations of the circles in standard form
The equations of the circles are:
1. \( C_1: x^2 + y^2 - 4x - 6y + 5 = 0 \)
2. \( C_2: x^2 + y^2 - 2x - 4y - 1 = 0 \)
3. \( C_3: x^2 + y^2 - 6x - 2y = 0 \)

### Step 2: Find the radical axis of the first two circles \( C_1 \) and \( C_2 \)
The radical axis can be found by subtracting the equations of the circles:
\[ C_1 - C_2 = 0 \]
This gives us:
\[ (x^2 + y^2 - 4x - 6y + 5) - (x^2 + y^2 - 2x - 4y - 1) = 0 \]
Simplifying this:
\[ -4x + 2x - 6y + 4y + 5 + 1 = 0 \]
\[ -2x - 2y + 6 = 0 \]
Dividing through by -2:
\[ x + y - 3 = 0 \]
Thus, the radical axis \( R_{12} \) is:
\[ x + y = 3 \quad \text{(Equation 1)} \]

### Step 3: Find the radical axis of the second and third circles \( C_2 \) and \( C_3 \)
Now we find the radical axis for circles \( C_2 \) and \( C_3 \):
\[ C_2 - C_3 = 0 \]
This gives us:
\[ (x^2 + y^2 - 2x - 4y - 1) - (x^2 + y^2 - 6x - 2y) = 0 \]
Simplifying this:
\[ -2x + 6x - 4y + 2y - 1 = 0 \]
\[ 4x - 2y - 1 = 0 \]
Rearranging gives:
\[ 4x - 2y = 1 \]
Dividing through by 2:
\[ 2x - y = \frac{1}{2} \quad \text{(Equation 2)} \]

### Step 4: Solve the system of equations formed by the radical axes
We now have two equations:
1. \( x + y = 3 \) (Equation 1)
2. \( 2x - y = \frac{1}{2} \) (Equation 2)

We can solve these equations simultaneously. From Equation 1, we can express \( y \) in terms of \( x \):
\[ y = 3 - x \]
Substituting this into Equation 2:
\[ 2x - (3 - x) = \frac{1}{2} \]
This simplifies to:
\[ 2x - 3 + x = \frac{1}{2} \]
\[ 3x - 3 = \frac{1}{2} \]
Adding 3 to both sides:
\[ 3x = \frac{1}{2} + 3 = \frac{1}{2} + \frac{6}{2} = \frac{7}{2} \]
Dividing by 3:
\[ x = \frac{7}{6} \]

### Step 5: Substitute back to find \( y \)
Now substituting \( x \) back into Equation 1 to find \( y \):
\[ y = 3 - \frac{7}{6} = \frac{18}{6} - \frac{7}{6} = \frac{11}{6} \]

### Step 6: Write the coordinates of the radical center
Thus, the radical center of the three circles is:
\[ \left( \frac{7}{6}, \frac{11}{6} \right) \]

### Summary of the Solution
The radical center of the circles is:
\[ \boxed{\left( \frac{7}{6}, \frac{11}{6} \right)} \]