To solve the problem, we need to show that four common tangents can be drawn for the given circles and find the internal and external centers of similitude.
### Step 1: Rewrite the equations of the circles in standard form
The equations of the circles are given as:
1. \( x^2 + y^2 - 14x + 6y + 33 = 0 \)
2. \( x^2 + y^2 + 30x - 2y + 1 = 0 \)
We will complete the square for both equations.
**Circle 1:**
\[
x^2 - 14x + y^2 + 6y + 33 = 0
\]
Completing the square for \(x\):
\[
x^2 - 14x = (x - 7)^2 - 49
\]
Completing the square for \(y\):
\[
y^2 + 6y = (y + 3)^2 - 9
\]
Substituting back:
\[
(x - 7)^2 - 49 + (y + 3)^2 - 9 + 33 = 0
\]
\[
(x - 7)^2 + (y + 3)^2 - 25 = 0
\]
Thus, the first circle is:
\[
(x - 7)^2 + (y + 3)^2 = 25
\]
This gives us center \(C_1(7, -3)\) and radius \(r_1 = 5\).
**Circle 2:**
\[
x^2 + 30x + y^2 - 2y + 1 = 0
\]
Completing the square for \(x\):
\[
x^2 + 30x = (x + 15)^2 - 225
\]
Completing the square for \(y\):
\[
y^2 - 2y = (y - 1)^2 - 1
\]
Substituting back:
\[
(x + 15)^2 - 225 + (y - 1)^2 - 1 + 1 = 0
\]
\[
(x + 15)^2 + (y - 1)^2 - 225 = 0
\]
Thus, the second circle is:
\[
(x + 15)^2 + (y - 1)^2 = 225
\]
This gives us center \(C_2(-15, 1)\) and radius \(r_2 = 15\).
### Step 2: Check the condition for four common tangents
The condition for two circles to have four common tangents is:
\[
d > r_1 + r_2
\]
where \(d\) is the distance between the centers of the circles.
**Finding the distance \(d\):**
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(-15 - 7)^2 + (1 + 3)^2}
\]
\[
= \sqrt{(-22)^2 + (4)^2} = \sqrt{484 + 16} = \sqrt{500} = 10\sqrt{5}
\]
**Calculating \(r_1 + r_2\):**
\[
r_1 + r_2 = 5 + 15 = 20
\]
**Comparing \(d\) and \(r_1 + r_2\):**
We need to check if \(10\sqrt{5} > 20\):
\[
10\sqrt{5} \approx 22.36 > 20
\]
Thus, the condition is satisfied, and we can conclude that there are four common tangents.
### Step 3: Find the internal and external centers of similitude
**Internal center of similitude \(O\):**
The internal center divides the line segment joining \(C_1\) and \(C_2\) in the ratio \(r_1:r_2\).
Using the formula:
\[
O = \left( \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}, \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} \right)
\]
where \(m_1 = r_1 = 5\), \(m_2 = r_2 = 15\), \(C_1(7, -3)\), and \(C_2(-15, 1)\):
\[
O = \left( \frac{5(-15) + 15(7)}{5 + 15}, \frac{5(1) + 15(-3)}{5 + 15} \right)
\]
\[
= \left( \frac{-75 + 105}{20}, \frac{5 - 45}{20} \right) = \left( \frac{30}{20}, \frac{-40}{20} \right) = \left( \frac{3}{2}, -2 \right)
\]
**External center of similitude \(O'\):**
The external center divides the line segment joining \(C_1\) and \(C_2\) in the ratio \(r_1:-r_2\).
Using the formula:
\[
O' = \left( \frac{m_1 x_2 - m_2 x_1}{m_1 - m_2}, \frac{m_1 y_2 - m_2 y_1}{m_1 - m_2} \right)
\]
\[
O' = \left( \frac{5(-15) - 15(7)}{5 - 15}, \frac{5(1) - 15(-3)}{5 - 15} \right)
\]
\[
= \left( \frac{-75 - 105}{-10}, \frac{5 + 45}{-10} \right) = \left( \frac{-180}{-10}, \frac{50}{-10} \right) = (18, -5)
\]
### Final Results:
- The four common tangents can be drawn for the circles.
- The internal center of similitude is \(O\left(\frac{3}{2}, -2\right)\).
- The external center of similitude is \(O'(18, -5)\).
