Find k if the following pairs of circles are orthogonal
`x^2+y^2+2by-k=0`
`x^2+y^2+2ax+8=0`

Answer

To find the value of \( k \) for which the given pairs of circles are orthogonal, we will follow these steps:

### Step 1: Write down the equations of the circles
The equations of the circles are given as:
1. \( S_1: x^2 + y^2 + 2by - k = 0 \)
2. \( S_2: x^2 + y^2 + 2ax + 8 = 0 \)

### Step 2: Identify the coefficients
We can rewrite the equations in the standard form \( x^2 + y^2 + 2gx + 2fy + c = 0 \) to identify the coefficients:
- For \( S_1 \):
- \( g_1 = 0 \)
- \( f_1 = b \)
- \( c_1 = -k \)

- For \( S_2 \):
- \( g_2 = a \)
- \( f_2 = 0 \)
- \( c_2 = 8 \)

### Step 3: Use the orthogonality condition
The circles are orthogonal if the following condition holds:
\[ 2f_1f_2 + 2g_1g_2 = c_1 + c_2 \]
Substituting the values we identified:
\[ 2(b)(0) + 2(0)(a) = -k + 8 \]
This simplifies to:
\[ 0 = -k + 8 \]

### Step 4: Solve for \( k \)
Rearranging the equation gives:
\[ k = 8 \]

### Final Answer
Thus, the value of \( k \) is:
\[ \boxed{8} \]