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 values of p and q respectively are

3, 1

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2, 2

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3, 2

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3, 4

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Solution

The correct option is C 3, 2
In the equation of circle, there is no term containing xy and coefficient of $x^{2}$ and $y^{2}$ are equal. Therefore 2 - q = 0 $\Rightarrow$ q = 2 and p = 3.

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