The value of `lambda` for which the circle `x^(2)+y^(2)+2lambdax+6y+1=0` intersects the circle `x^(2)+y^(2)+4x+2y=0` orthogonally, is

Options

A. `(11)/(8)`
B. -1
C. `(-5)/(4)`
D. `(5)/(2)`

Answer

To find the value of \( \lambda \) for which the circles intersect orthogonally, we will use the condition for orthogonality of two circles.

### Step-by-Step Solution:

1. **Identify the equations of the circles:**
The first circle is given by:
\[ x^2 + y^2 + 2\lambda x + 6y + 1 = 0 \]
The second circle is given by:
\[ x^2 + y^2 + 4x + 2y = 0 \]

2. **Rewrite the equations in standard form:**
For the first circle, we can identify:
- \( 2g = 2\lambda \) → \( g = \lambda \)
- \( 2f = 6 \) → \( f = 3 \)
- \( c = 1 \)

For the second circle:
- \( 2g_1 = 4 \) → \( g_1 = 2 \)
- \( 2f_1 = 2 \) → \( f_1 = 1 \)
- \( c_1 = 0 \)

3. **Apply the orthogonality condition:**
The condition for two circles to intersect orthogonally is given by:
\[ 2gg_1 + 2ff_1 = c + c_1 \]
Substituting the values we found:
\[ 2(\lambda)(2) + 2(3)(1) = 1 + 0 \]
Simplifying this gives:
\[ 4\lambda + 6 = 1 \]

4. **Solve for \( \lambda \):**
Rearranging the equation:
\[ 4\lambda = 1 - 6 \]
\[ 4\lambda = -5 \]
\[ \lambda = -\frac{5}{4} \]

Thus, the value of \( \lambda \) for which the circles intersect orthogonally is:
\[ \lambda = -\frac{5}{4} \]