The internal centre of similitude of the two circles `x^(2)+y^(2)+6x-2y+1=0, x^(2)+y^(2)-2x-6y+9=0` is

Answer

To find the internal center of similitude of the two circles given by the equations:

1. \( x^2 + y^2 + 6x - 2y + 1 = 0 \)
2. \( x^2 + y^2 - 2x - 6y + 9 = 0 \)

we will follow these steps:

### Step 1: Rewrite the equations in standard form
We need to rewrite both equations in the standard form of a circle, which is \((x - h)^2 + (y - k)^2 = r^2\).

**For the first circle:**
\[ x^2 + y^2 + 6x - 2y + 1 = 0 \]
Rearranging gives:
\[ x^2 + 6x + y^2 - 2y = -1 \]
Completing the square for \(x\) and \(y\):
\[ (x^2 + 6x + 9) + (y^2 - 2y + 1) = -1 + 9 + 1 \]
\[ (x + 3)^2 + (y - 1)^2 = 9 \]
This gives us the center \(C_1(-3, 1)\) and radius \(r_1 = 3\).

**For the second circle:**
\[ x^2 + y^2 - 2x - 6y + 9 = 0 \]
Rearranging gives:
\[ x^2 - 2x + y^2 - 6y = -9 \]
Completing the square for \(x\) and \(y\):
\[ (x^2 - 2x + 1) + (y^2 - 6y + 9) = -9 + 1 + 9 \]
\[ (x - 1)^2 + (y - 3)^2 = 1 \]
This gives us the center \(C_2(1, 3)\) and radius \(r_2 = 1\).

### Step 2: Find the internal center of similitude
The formula for the internal center of similitude \(S_i\) is given by:
\[ S_i = \left( \frac{r_1 \cdot x_2 + r_2 \cdot x_1}{r_1 + r_2}, \frac{r_1 \cdot y_2 + r_2 \cdot y_1}{r_1 + r_2} \right) \]
Where \((x_1, y_1)\) and \((x_2, y_2)\) are the centers of the circles.

Substituting the values:
- \(C_1(-3, 1)\) gives \(x_1 = -3\), \(y_1 = 1\)
- \(C_2(1, 3)\) gives \(x_2 = 1\), \(y_2 = 3\)
- \(r_1 = 3\) and \(r_2 = 1\)

Now substituting into the formula:
\[ S_i = \left( \frac{3 \cdot 1 + 1 \cdot (-3)}{3 + 1}, \frac{3 \cdot 3 + 1 \cdot 1}{3 + 1} \right) \]
Calculating the x-coordinate:
\[ S_{ix} = \frac{3 - 3}{4} = \frac{0}{4} = 0 \]
Calculating the y-coordinate:
\[ S_{iy} = \frac{9 + 1}{4} = \frac{10}{4} = 2.5 \]

Thus, the internal center of similitude is:
\[ S_i = (0, 2.5) \]

### Final Answer
The internal center of similitude of the two circles is \( (0, 2.5) \).

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