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 = \(\sqrt{g^2+f^2-c}\)
= \(\sqrt{1^2+(-\frac{2}{3})^2-(-\frac{1}{3})}\)
= \(\sqrt{1+\frac{4}{9}+\frac{1}{3}}\) = \(\sqrt{\frac{16}{9}}\) = \(\frac{4}{3}\)