consider two curves `ax^2+4xy+2y^2+x+y+5=0` and `ax^2+6xy+5y^2+2x+3y+8=0` these two curves intersect at four cocyclic points then find out `a`

Options

A. `-6`
B. `-4`
C. 4
D. 6

Answer

To solve the problem of finding the value of \( a \) such that the curves \( ax^2 + 4xy + 2y^2 + x + y + 5 = 0 \) and \( ax^2 + 6xy + 5y^2 + 2x + 3y + 8 = 0 \) intersect at four cocyclic points, we can follow these steps:

### Step 1: Define the curves
Let:
- \( S_1 = ax^2 + 4xy + 2y^2 + x + y + 5 = 0 \)
- \( S_2 = ax^2 + 6xy + 5y^2 + 2x + 3y + 8 = 0 \)

### Step 2: Set up the family of curves
Since the curves intersect at four cocyclic points, we can form a family of curves by taking a linear combination of \( S_1 \) and \( S_2 \):
\[ S_1 - \lambda S_2 = 0 \]
This gives us:
\[ (ax^2 + 4xy + 2y^2 + x + y + 5) - \lambda(ax^2 + 6xy + 5y^2 + 2x + 3y + 8) = 0 \]

### Step 3: Expand and collect like terms
Expanding this equation, we have:
\[ (a - \lambda a)x^2 + (4 - 6\lambda)xy + (2 - 5\lambda)y^2 + (1 - 2\lambda)x + (1 - 3\lambda)y + (5 - 8\lambda) = 0 \]

### Step 4: Identify conditions for a circle
For this equation to represent a circle, the following conditions must hold:
1. Coefficient of \( x^2 \) must equal the coefficient of \( y^2 \):
\[ a - \lambda a = 2 - 5\lambda \]
2. Coefficient of \( xy \) must be zero:
\[ 4 - 6\lambda = 0 \]

### Step 5: Solve for \( \lambda \)
From the second condition:
\[ 4 - 6\lambda = 0 \implies \lambda = \frac{4}{6} = \frac{2}{3} \]

### Step 6: Substitute \( \lambda \) back into the first condition
Substituting \( \lambda = \frac{2}{3} \) into the first condition:
\[ a - \frac{2}{3}a = 2 - 5 \cdot \frac{2}{3} \]
This simplifies to:
\[ \frac{1}{3}a = 2 - \frac{10}{3} \]
\[ \frac{1}{3}a = \frac{6}{3} - \frac{10}{3} = -\frac{4}{3} \]

### Step 7: Solve for \( a \)
Multiplying both sides by 3 gives:
\[ a = -4 \]

### Conclusion
Thus, the value of \( a \) is:
\[ \boxed{-4} \]