Show that the equation $3 x^{2} + 3 y^{2} - 6 x + 4 y - 1 = 0$ represents a circle. Find its centre and radius.

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Solution

$3 x^{2} + 3 y^{2} - 6 x + 4 y - 1 = 0 \Rightarrow [(\sqrt{3} x)^{2} - 2 \times \sqrt{3} \times \sqrt{3} \times x + (\sqrt{3})^{2}] + [(\sqrt{3} y)^{2} + 2 \times \sqrt{3} \times \frac{2}{\sqrt{3}} y + (\frac{2}{\sqrt{3}})^{2}] - 3 - \frac{4}{3} - 1 = 0$
$\Rightarrow (\sqrt{3})^{2} + (\sqrt{3} y + \frac{2}{\sqrt{3}})^{2} - \frac{16}{3} = 0$
$3 (x - 1)^{2} + 3 (y + \frac{2}{3})^{2} - \frac{16}{3} = 0$
$(x - 1)^{2} + (y + \frac{2}{3})^{2} = \frac{16}{3 \times 3} = (\frac{4}{3})^{2}$
center $(1, \frac{- 2}{3})$ radius = $\frac{4}{3}$

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Show that the equation 3x2+3y2−6x+4y−1=0 represents a circle. Find its centre and radius.

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Show that the equation $3 x^{2} + 3 y^{2} - 6 x + 4 y - 1 = 0$ represents a circle. Find its centre and radius.