If the two circles `x^(2)+y^(2)-2x+22y+5=0andx^(2)+y^(2)+14x+6y+k=0` intersect orthogonally ,k is equal to

Options

A. 47
B. `-47`
C. 49
D. `-49`

Answer

To solve the problem of finding the value of \( k \) such that the two circles intersect orthogonally, we will follow these steps:

### Step 1: Identify the coefficients from the equations of the circles

The equations of the circles are given as:
1. \( x^2 + y^2 - 2x + 22y + 5 = 0 \)
2. \( x^2 + y^2 + 14x + 6y + k = 0 \)

From the general form of a circle \( x^2 + y^2 + 2gx + 2fy + c = 0 \), we can identify the coefficients:
- For the first circle:
- \( 2g_1 = -2 \) → \( g_1 = -1 \)
- \( 2f_1 = 22 \) → \( f_1 = 11 \)
- \( c_1 = 5 \)

- For the second circle:
- \( 2g_2 = 14 \) → \( g_2 = 7 \)
- \( 2f_2 = 6 \) → \( f_2 = 3 \)
- \( c_2 = k \)

### Step 2: Use the condition for orthogonal intersection of circles

The condition for two circles to intersect orthogonally is given by:
\[ 2(g_1g_2 + f_1f_2) = c_1 + c_2 \]

### Step 3: Substitute the values into the condition

Substituting the values we found:
- \( g_1 = -1 \)
- \( g_2 = 7 \)
- \( f_1 = 11 \)
- \( f_2 = 3 \)
- \( c_1 = 5 \)
- \( c_2 = k \)

We can substitute these into the orthogonality condition:
\[ 2((-1)(7) + (11)(3)) = 5 + k \]

### Step 4: Calculate the left-hand side

Calculating the left-hand side:
\[ 2(-7 + 33) = 2(26) = 52 \]

### Step 5: Set up the equation

Now we set up the equation:
\[ 52 = 5 + k \]

### Step 6: Solve for \( k \)

To find \( k \), we rearrange the equation:
\[ k = 52 - 5 = 47 \]

### Final Answer

Thus, the value of \( k \) is:
\[ \boxed{47} \]