Find k if the following pairs of circles are orthogonal
`x^2+y^2-6x-8y+12=0`
`x^2+y^2-4x+6y+k=0`

Answer

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

### Step 1: Write the equations of the circles
The equations of the circles are given as:
1. \( x^2 + y^2 - 6x - 8y + 12 = 0 \)
2. \( x^2 + y^2 - 4x + 6y + k = 0 \)

### Step 2: Compare with the general form of a circle
The general form of a circle is:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]

For the first circle:
- Comparing, we have:
- \( 2g_1 = -6 \) → \( g_1 = -3 \)
- \( 2f_1 = -8 \) → \( f_1 = -4 \)
- \( c_1 = 12 \)

For the second circle:
- Comparing, we have:
- \( 2g_2 = -4 \) → \( g_2 = -2 \)
- \( 2f_2 = 6 \) → \( f_2 = 3 \)
- \( c_2 = k \)

### Step 3: Use the condition for orthogonality
The circles are orthogonal if:
\[ 2g_1g_2 + 2f_1f_2 = c_1 + c_2 \]

Substituting the values:
- \( g_1 = -3 \)
- \( g_2 = -2 \)
- \( f_1 = -4 \)
- \( f_2 = 3 \)
- \( c_1 = 12 \)
- \( c_2 = k \)

We substitute these values into the orthogonality condition:
\[ 2(-3)(-2) + 2(-4)(3) = 12 + k \]

### Step 4: Simplify the equation
Calculating the left side:
\[ 2(6) + 2(-12) = 12 + k \]
\[ 12 - 24 = 12 + k \]
\[ -12 = 12 + k \]

### Step 5: Solve for \( k \)
Now, we isolate \( k \):
\[ k = -12 - 12 \]
\[ k = -24 \]

### Conclusion
Thus, the value of \( k \) is \( -24 \).

---