Problem: If the curves $ax^2 + 4xy + 2y^2 + x+ y + 5 = 0 $ and $ax^2 + 6xy + 5y^2 + 2x + 3y + 8 = 0$ intersect at four concyclic points then the value of a is
(A) 4
(B) -4
(C) 6
(D) –6
Solution: Any second degree curve passing through the intersection of the given curves is
ax2 + 4xy + 2y2 + x + y + 5 + λ ( ax2 + 6xy + 5y2 +2 x + 3y + 8 ) = 0
If it is a circle, then coefficient of x2 = coefficient of y2 and coefficient of xy = 0
a(1+ λ) = 2 + 5λ and 4 + 6λ = 0
$\large a = \frac{2+5\lambda}{1+\lambda} $ and $\lambda =- \frac{2}{3}$
$\large a = \frac{2-\frac{10}{3}}{1-\frac{2}{3}} = -4$
Hence (B) is the correct answer.