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Family of Circles

If the curves...

Question

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

A

4

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B

-4

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C

6

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D

-6

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Solution

The correct option is B -4
$(a x^{2} + 4 x y + 2 y^{2} + x + y + 5) + λ (a x^{2} + 6 x y + 5 y^{2} + 2 x + 3 y + 8) = 0$
Coefficient $x^{2}$ = Coefficient $y^{2}$
a $(1 + λ) = 2 + 5 λ$
Coefficient xy = 0
$4 + 6 λ = 0 \Rightarrow λ = \frac{- 2}{3}$

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