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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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Answer» 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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