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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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The general equation of a conic is as follows 

ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 where a, b, c, f, g, h are constants 

For a circle, a = b and h = 0. 

The equation becomes: 

x2 + y2 + 2gx + 2fy + c = 0…(i)

Given, 3x2 + 3y2 + 6x - 4y - 1

⇒ x2 + y2 + 2x - \(\frac{4}{3}\)y - \(\frac{1}{3}\) = 0

Comparing with (i) we see that the equation represents a circle with 2g = 2 

⇒ g = 1.2f = -\(\frac{4}{3}\) 

⇒ f =  -\(\frac{2}{3}\) and c =  -\(\frac{1}{3}\) 

Centre (-g,-f) = {-1,-(-\(\frac{2}{3}\))}

\(\Big(-1,\frac{2}{3}\Big)\) 

Radius = \(\sqrt{g^2+f^2-c}\) 

=   \(\sqrt{1^2+(-\frac{2}{3})^2-(-\frac{1}{3})}\) 

\(\sqrt{1+\frac{4}{9}+\frac{1}{3}}\) = \(\sqrt{\frac{16}{9}}\) = \(\frac{4}{3}\)

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