If the equation $p x^{2} + (2 - q) x y + 3 y^{2} - 6 q x + 30 y + 6 q = 0$ represents a circle, then the value of $p \times q$ is

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Solution

Given $p x^{2} + (2 - q) x y + 3 y^{2} - 6 q x + 30 y + 6 q = 0$ represents a circle, so

Coefficient of $x y$ term is $0$
$2 - q = 0 \Rightarrow q = 2$
Coefficient of $x^{2}$ and $y^{2}$ are equal
$p = 3$
Now,
$3 x^{2} + 3 y^{2} - 12 x + 30 y + 12 = 0 \Rightarrow x^{2} + y^{2} - 4 x + 10 y + 4 = 0 \Rightarrow (x - 2)^{2} + (y + 5)^{2} = 25$
Which is a circle

Hence, $p \times q = 6$

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