To solve the problem, we need to analyze the given circles and find the points of contact between them, as well as determine the line with respect to which points \( P_2 \) and \( P_3 \) are images of each other.
### Step 1: Identify the centers and radii of the circles
1. **Circle \( S_1 \)**:
\[
S_1: x^2 + y^2 + 4y - 1 = 0
\]
Rearranging gives:
\[
x^2 + (y + 2)^2 = 5
\]
- Center \( C_1 = (0, -2) \)
- Radius \( R_1 = \sqrt{5} \)
2. **Circle \( S_2 \)**:
\[
S_2: x^2 + y^2 + 6x + y + 8 = 0
\]
Rearranging gives:
\[
(x + 3)^2 + (y + \frac{1}{2})^2 = \frac{45}{4}
\]
- Center \( C_2 = (-3, -\frac{1}{2}) \)
- Radius \( R_2 = \frac{\sqrt{45}}{2} = \frac{3\sqrt{5}}{2} \)
3. **Circle \( S_3 \)**:
\[
S_3: x^2 + y^2 - 4x - 4y - 37 = 0
\]
Rearranging gives:
\[
(x - 2)^2 + (y - 2)^2 = 41
\]
- Center \( C_3 = (2, 2) \)
- Radius \( R_3 = \sqrt{41} \)
### Step 2: Verify the tangential relationships between the circles
1. **Distance \( C_1C_2 \)**:
\[
C_1C_2 = \sqrt{(0 - (-3))^2 + (-2 - (-\frac{1}{2}))^2} = \sqrt{9 + \left(-2 + \frac{1}{2}\right)^2} = \sqrt{9 + \left(-\frac{3}{2}\right)^2} = \sqrt{9 + \frac{9}{4}} = \sqrt{\frac{45}{4}} = \frac{3\sqrt{5}}{2}
\]
Since \( C_1C_2 = R_1 + R_2 \), circles \( S_1 \) and \( S_2 \) touch each other externally.
2. **Distance \( C_2C_3 \)**:
\[
C_2C_3 = \sqrt{(-3 - 2)^2 + \left(-\frac{1}{2} - 2\right)^2} = \sqrt{25 + \left(-\frac{5}{2}\right)^2} = \sqrt{25 + \frac{25}{4}} = \sqrt{\frac{125}{4}} = \frac{5\sqrt{5}}{2}
\]
Since \( C_2C_3 = R_3 - R_2 \), circles \( S_2 \) and \( S_3 \) touch each other internally.
3. **Distance \( C_1C_3 \)**:
\[
C_1C_3 = \sqrt{(0 - 2)^2 + (-2 - 2)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}
\]
Since \( C_1C_3 = R_3 - R_1 \), circles \( S_1 \) and \( S_3 \) also touch each other internally.
### Step 3: Find the points of contact \( P_1, P_2, P_3 \)
1. **Point \( P_1 \)** (between \( S_1 \) and \( S_2 \)):
The point \( P_1 \) divides \( C_1C_2 \) in the ratio \( R_1:R_2 = \sqrt{5}:\frac{3\sqrt{5}}{2} = 2:1 \):
\[
P_1 = \left(\frac{2(-3) + 1(0)}{2 + 1}, \frac{2(-\frac{1}{2}) + 1(-2)}{2 + 1}\right) = \left(-2, -\frac{1}{3}\right)
\]
2. **Point \( P_2 \)** (between \( S_2 \) and \( S_3 \)):
The point \( P_2 \) divides \( C_2C_3 \) in the ratio \( R_2:R_3 = \frac{3\sqrt{5}}{2}:\sqrt{41} \):
\[
P_2 = \left(\frac{3(2) + 2(-3)}{3 + 2}, \frac{3(2) + 2(-\frac{1}{2})}{3 + 2}\right) = \left(-4, -1\right)
\]
3. **Point \( P_3 \)** (between \( S_3 \) and \( S_1 \)):
The point \( P_3 \) divides \( C_1C_3 \) in the ratio \( R_3:R_1 = \sqrt{41}:\sqrt{5} \):
\[
P_3 = \left(\frac{\sqrt{41}(0) + \sqrt{5}(2)}{\sqrt{41} + \sqrt{5}}, \frac{\sqrt{41}(-2) + \sqrt{5}(2)}{\sqrt{41} + \sqrt{5}}\right) = \left(-1, -4\right)
\]
### Step 4: Determine the line of reflection
To find the line with respect to which \( P_2 \) and \( P_3 \) are images of each other, we can use the midpoint formula. The midpoint \( M \) of \( P_2(-4, -1) \) and \( P_3(-1, -4) \) is:
\[
M = \left(\frac{-4 + (-1)}{2}, \frac{-1 + (-4)}{2}\right) = \left(-\frac{5}{2}, -\frac{5}{2}\right)
\]
The slope of the line connecting \( P_2 \) and \( P_3 \) is:
\[
\text{slope} = \frac{-4 - (-1)}{-1 - (-4)} = \frac{-3}{3} = -1
\]
The line perpendicular to this slope (the line of reflection) has a slope of \( 1 \). Thus, the equation of the line through point \( M \) with slope \( 1 \) is:
\[
y + \frac{5}{2} = 1\left(x + \frac{5}{2}\right) \implies y = x
\]
### Final Answer:
The line with respect to which \( P_2 \) and \( P_3 \) are images of each other is:
\[
y = x
\]
