Find k if the following pairs of circles are orthogonal
`x^2+y^2-16y-k=0`
`x^2+y^2+4x-8=0`

Answer

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

### Step 1: Write the equations of the circles in standard form
The equations of the circles given are:
1. \( x^2 + y^2 - 16y - k = 0 \)
2. \( x^2 + y^2 + 4x - 8 = 0 \)

### Step 2: Identify the coefficients from the general form of the circle
The general form of a circle is given by:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
From the first circle:
- Comparing, we have \( g_1 = 0 \), \( f_1 = -8 \), and \( c_1 = -k \).

From the second circle:
- Comparing, we have \( g_2 = 2 \), \( f_2 = 0 \), and \( c_2 = -8 \).

### Step 3: Use the orthogonality condition
For two circles to be orthogonal, the following condition must hold:
\[ 2f_1f_2 + 2g_1g_2 = c_1 + c_2 \]

### Step 4: Substitute the values into the orthogonality condition
Substituting the values we found:
- \( f_1 = -8 \), \( f_2 = 0 \)
- \( g_1 = 0 \), \( g_2 = 2 \)
- \( c_1 = -k \), \( c_2 = -8 \)

So we have:
\[ 2(-8)(0) + 2(0)(2) = -k - 8 \]

### Step 5: Simplify the equation
This simplifies to:
\[ 0 + 0 = -k - 8 \]
Thus:
\[ 0 = -k - 8 \]

### Step 6: Solve for \( k \)
Rearranging gives:
\[ -k = 8 \]
So:
\[ k = -8 \]

### Final Answer
The value of \( k \) is \( -8 \).

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