If the equation $a x^{2} + (b - 3) x y + 3 y^{2} + 6 a x + 2 b y - 3 = 0$ represents a circle, then the value of $a^{2} + b^{2}$ is

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Solution

Given equation:
$a x^{2} + (b - 3) x y + 3 y^{2} + 6 a x + 2 b y - 3 = 0$
This represent circle iff
$(i)$ Coefficient of $x^{2}$ =Coefficient of $y^{2}$
$\Rightarrow a = 3$
$(i i)$ Coefficient of $x y = 0$
$\Rightarrow b - 3 = 0 \Rightarrow b = 3$
$∴ a^{2} + b^{2} = 18$

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