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

Answer 1

Compute the required value using condition on coefficients

We know that, any second degree curve that passes through the intersection of the given curves is $a{x}^2+4xy+2y^2+x+y+5+\lambda (a{x}^2+6xy+5y^2+2x+3y+8)=0$

So, if it is a circle, then coefficient of $x^2$$=$coefficient of $y^2$ and the coefficient of $xy=0$.

coefficient of $x^2$$=$coefficient of $y^2$

$$\begin{gathered} a(1+\lambda )=2+5\lambda \\ \Rightarrow a=\frac{2+5\lambda }{(1+\lambda )} \end{gathered}$$ ……………………$1$

coefficient of $xy=0$

$$\begin{gathered} 4+6\lambda =0 \\ \Rightarrow \lambda =-\frac{4}{6} \end{gathered}$$

Substitute the value of $\lambda$ in equation $1$

$$\begin{gathered} a=\frac{2+5\left(-{\displaystyle \frac{4}{6}}\right)}{1+\left(-{\displaystyle \frac{4}{6}}\right)} \\ \Rightarrow a=\frac{2-{\displaystyle \frac{10}{3}}}{1-{\displaystyle \frac{2}{3}}} \\ \Rightarrow a=\frac{6-10}{3-2} \\ \Rightarrow a=-4 \end{gathered}$$

Hence, the value of $a$ is $-4$.