If a line L touches the circle `x^2 + y^2 + 8x + 15 = 0` at A in second quadrant and circle `x^2 + y^2 - 12x + 32 = 0` at B in fourth quadrant and d is distance between A and B then `d^2` is:

Explanation

To solve the problem, we need to find the distance squared \( d^2 \) between the points \( A \) and \( B \) where the line \( L \) touches the two given circles. Let's break down the solution step by step.

### Step 1: Rewrite the equations of the circles in standard form

1. **Circle 1**:
\[ x^2 + y^2 + 8x + 15 = 0 \]
Rearranging gives:
\[ x^2 + 8x + y^2 + 15 = 0 \]
Completing the square for \( x \):
\[ (x + 4)^2 - 16 + y^2 + 15 = 0 \implies (x + 4)^2 + y^2 - 1 = 0 \implies (x + 4)^2 + y^2 = 1 \]
This gives us the center \( C_1(-4, 0) \) and radius \( r_1 = 1 \).

2. **Circle 2**:
\[ x^2 + y^2 - 12x + 32 = 0 \]
Rearranging gives:
\[ x^2 - 12x + y^2 + 32 = 0 \]
Completing the square for \( x \):
\[ (x - 6)^2 - 36 + y^2 + 32 = 0 \implies (x - 6)^2 + y^2 - 4 = 0 \implies (x - 6)^2 + y^2 = 4 \]
This gives us the center \( C_2(6, 0) \) and radius \( r_2 = 2 \).

### Step 2: Determine the coordinates of points A and B

- **Point A** lies in the second quadrant on Circle 1. The coordinates can be represented as:
\[ A(-4, y_A) \quad \text{where } y_A = \sqrt{1 - (x + 4)^2} = \sqrt{1 - 0} = 1 \implies A(-4, 1) \]

- **Point B** lies in the fourth quadrant on Circle 2. The coordinates can be represented as:
\[ B(6, y_B) \quad \text{where } y_B = -\sqrt{4 - (x - 6)^2} = -\sqrt{4 - 0} = -2 \implies B(6, -2) \]

### Step 3: Calculate the distance \( d \) between points A and B

Using the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the coordinates of points A and B:
\[ d = \sqrt{(6 - (-4))^2 + (-2 - 1)^2} = \sqrt{(6 + 4)^2 + (-3)^2} = \sqrt{10^2 + (-3)^2} = \sqrt{100 + 9} = \sqrt{109} \]

### Step 4: Calculate \( d^2 \)

Now, we find \( d^2 \):
\[ d^2 = 109 \]

### Final Answer
Thus, the value of \( d^2 \) is:
\[ \boxed{109} \]