Question

If the curves $a{x}^2+4xy+2y^2+x+y+5=0$ and $a{x}^2+6xy+5y^2+2x+3y+8=0$ intersect at four con-cyclic points then the value of $a$ is

• A. $4$
• B. $–4$
• C. $6$
• D. $–6$

Answer 1

The correct option is B $–4$

Any second degree curve passing through the intersection of the given curves is

$a{x}^2+4xy+2y^2+x+y+5+\lambda (a{x}^2+6xy+5y^2+2x+3y+8)=0$

If it is a circle, then coefficient of $x^2=$ coefficient of $y^2$ and coefficient of $xy=0$

$\Rightarrow a(1+\lambda )=2+5\lambda$ and $4+6\lambda =0$

$\Rightarrow a=\frac{2+5\lambda }{1+\lambda }$ and $\lambda =\frac{-2}{3}$

$\Rightarrow a=\frac{2+5\times \frac{-2}{3}}{1-\frac{2}{3}}=\frac{2-\frac{10}{3}}{1-\frac{2}{3}}=\frac{6-10}{3-2}=-4$