Key Concepts and Formulas

• The standard equation of a circle with center $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$.
• The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
• The number of common tangents between two circles depends on the relationship between the distance between their centers ($d$) and their radii ($r_1$ and $r_2$). If $d > r_1 + r_2$, there are 4 common tangents.

Step-by-Step Solution

1. Convert Circle Equations to Standard Form

We need to find the center and radius of each circle. Converting the given equations to standard form $(x-h)^2 + (y-k)^2 = r^2$ will allow us to easily identify these.

Circle 1: $x^2 + y^2 - 18x - 15y + 131 = 0$ Completing the square for $x$ and $y$ terms: $(x^2 - 18x) + (y^2 - 15y) = -131$ $(x^2 - 18x + 81) + (y^2 - 15y + \frac{225}{4}) = -131 + 81 + \frac{225}{4}$ $(x - 9)^2 + (y - \frac{15}{2})^2 = -50 + \frac{225}{4}$ $(x - 9)^2 + (y - \frac{15}{2})^2 = \frac{-200 + 225}{4}$ $(x - 9)^2 + (y - \frac{15}{2})^2 = \frac{25}{4}$ Center $C_1 = (9, \frac{15}{2})$ and radius $r_1 = \sqrt{\frac{25}{4}} = \frac{5}{2}$.

Circle 2: $x^2 + y^2 - 6x - 6y - 7 = 0$ Completing the square for $x$ and $y$ terms: $(x^2 - 6x) + (y^2 - 6y) = 7$ $(x^2 - 6x + 9) + (y^2 - 6y + 9) = 7 + 9 + 9$ $(x - 3)^2 + (y - 3)^2 = 25$ Center $C_2 = (3, 3)$ and radius $r_2 = \sqrt{25} = 5$.

2. Calculate the Distance Between the Centers ($d$)

We use the distance formula to find the distance $d$ between the centers $C_1(9, \frac{15}{2})$ and $C_2(3, 3)$: $d = \sqrt{(9 - 3)^2 + (\frac{15}{2} - 3)^2}$ $d = \sqrt{(6)^2 + (\frac{15 - 6}{2})^2}$ $d = \sqrt{36 + (\frac{9}{2})^2}$ $d = \sqrt{36 + \frac{81}{4}}$ $d = \sqrt{\frac{144 + 81}{4}}$ $d = \sqrt{\frac{225}{4}}$ $d = \frac{15}{2}$

3. Compare $d$ with $r_1 + r_2$ and $|r_1 - r_2|$

We have $r_1 = \frac{5}{2}$, $r_2 = 5$, and $d = \frac{15}{2}$.

First, let's calculate $r_1 + r_2$: $r_1 + r_2 = \frac{5}{2} + 5 = \frac{5 + 10}{2} = \frac{15}{2}$ So, $d = r_1 + r_2 = \frac{15}{2}$. This means the circles touch externally.

Since the circles touch externally, there are 3 common tangents.

Common Mistakes & Tips

• Double-check your calculations when completing the square to avoid errors in determining the center and radius.
• Remember the different conditions for the number of common tangents based on the relationship between $d$, $r_1$, and $r_2$.
• Use the standard formulas for the center and radius from the general form of the circle equation to save time.

Summary

We first converted the equations of the two circles to standard form to find their centers and radii. Then, we calculated the distance between the centers and compared it to the sum of the radii. We found that the distance between the centers is equal to the sum of the radii, indicating that the circles touch externally. Therefore, there are 3 common tangents. Since the provided "Correct Answer" is 4, there must be a mistake in the "Correct Answer" provided. However, based on my calculations the number of common tangents should be 3.

Final Answer

The number of common tangents is 3, which corresponds to option (C).