The circle `x^(2)+y^(2) + 4x+6y - 8 = 0` and `x^(2)+y^(2) +6x-8y +c=0` cut orthogonally if c=

Options

A. 4
B. `-4`
C. 3
D. `-3`

Answer

To determine the value of \( c \) for which the circles \( x^2 + y^2 + 4x + 6y - 8 = 0 \) and \( x^2 + y^2 + 6x - 8y + c = 0 \) cut orthogonally, we will follow these steps:

### Step 1: Rewrite the equations of the circles
The equations of the circles can be rewritten in the standard form of a circle, which is given by:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
For the first circle:
\[ x^2 + y^2 + 4x + 6y - 8 = 0 \]
Here, we identify:
- \( g_1 = 2 \) (since \( 2g_1 = 4 \))
- \( f_1 = 3 \) (since \( 2f_1 = 6 \))
- \( c_1 = -8 \)

For the second circle:
\[ x^2 + y^2 + 6x - 8y + c = 0 \]
Here, we identify:
- \( g_2 = 3 \) (since \( 2g_2 = 6 \))
- \( f_2 = -4 \) (since \( 2f_2 = -8 \))
- \( c_2 = c \)

### Step 2: Use the orthogonality condition
Two circles cut orthogonally if the following condition holds:
\[ g_1 g_2 + f_1 f_2 = c_1 + c_2 \]

### Step 3: Substitute the values into the condition
Substituting the values we found:
\[ (2)(3) + (3)(-4) = -8 + c \]
Calculating the left side:
\[ 6 - 12 = -8 + c \]
This simplifies to:
\[ -6 = -8 + c \]

### Step 4: Solve for \( c \)
Now, we can solve for \( c \):
\[ c = -6 + 8 \]
\[ c = 2 \]

### Final Answer
Thus, the value of \( c \) for which the circles cut orthogonally is:
\[ \boxed{2} \]