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 \( x^2 + y^2 + 2\lambda x + 6y + 1 = 0 \) and \( x^2 + y^2 + 4x + 2y = 0 \) intersect orthogonally, we can follow these steps:

### Step 1: Identify the coefficients of the circles
The general form of a circle is given by:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]

For the first circle \( S_1: x^2 + y^2 + 2\lambda x + 6y + 1 = 0 \):
- \( g = \lambda \)
- \( f = 3 \)
- \( c = 1 \)

For the second circle \( S_2: x^2 + y^2 + 4x + 2y = 0 \):
- \( g_1 = 2 \)
- \( f_1 = 1 \)
- \( c_1 = 0 \)

### Step 2: Use the orthogonality condition
The condition for two circles to intersect orthogonally is given by:
\[ 2g \cdot g_1 + 2f \cdot f_1 = c + c_1 \]

Substituting the values we found:
\[ 2(\lambda)(2) + 2(3)(1) = 1 + 0 \]

### Step 3: Simplify the equation
This simplifies to:
\[ 4\lambda + 6 = 1 \]

### Step 4: Solve for \( \lambda \)
Now, we can isolate \( \lambda \):
\[ 4\lambda = 1 - 6 \]
\[ 4\lambda = -5 \]
\[ \lambda = -\frac{5}{4} \]

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