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 concyclic points then the value of $a$ is

• A. $4$
• B. $-4$
• C. $6$
• D. $-6$

Answer 1

Answer: (B)

$-4$

$S_1=a{x}^2+4xy+2y^2+x+y+5=0$

$S_2=a{x}^2+6xy+5y^2+2x+3y+8=0$

Equation of curve passing through their intersection will be $S_1+k{S}_2=0$

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

Since the points are concyclic, the above curve should be a circle ,

Thus,Coefficient of $xy=0$

$\Rightarrow 4+6k=0$

$\Rightarrow k=\frac{-2}{3}$

Ceofficient of $x^2=$Coefficient of $y^2$

$\Rightarrow (a+ka)=2+5k$

$\Rightarrow a(1+k)=2+5k$

$\Rightarrow a(1-\frac{2}{3})=2-\frac{10}{3}$

$\Rightarrow a\left(\frac{3-2}{3}\right)=\frac{6-10}{3}$

$\Rightarrow a=-4$