Find the coordinates of the centre of the circle 3x\(^2\) + 3y\(^2\) - 6x + 9y - 5 = 0

To find the center of the circle, we need to rewrite the equation in the standard form: (x - h)\(^2\) + (y - k)\(^2\) = r\(^2\) where (h,k) is the center of the circle and r is the radius. Starting with the given equation: 3x\(^2\) + 3y\(^2\) - 6x + 9y - 5 = 0 We can group the x and y terms together: (3x\(^2\) - 6x) + (3y\(^2\) + 9y) = 5 Next, we need to complete the square for both x and y. For the x terms, we can factor out a 3 from the first two terms: 3(x\(^2\) - 2x) To complete the square, we need to add and subtract (\(\frac{2}{2}\))\(^2\) = 1 inside the parenthesis: 3(x\(^2\) - 2x + 1 - 1) Then, we can simplify this expression: 3((x - 1)\(^2\) - 1) = 3(x - 1)\(^2\) - 3 For the y terms, we can follow the same process: 3(y\(^2\) + 3y) = 3(y\(^2\) + 3y + (\(\frac{3}{2}\))\(^2\) - (\(\frac{3}{2}\))\(^2\)) = 3((y + \(\frac{3}{2}\))\(^2\) - \(\frac{9}{4}\)) = 3(y + \(\frac{3}{2}\))\(^2\) - \(\frac{27}{4}\) Now we can substitute these expressions back into the original equation and simplify: 3(x - 1)\(^2\) - 3 + 3(y + \(\frac{3}{2}\))\(^2\) - \(\frac{27}{4}\) = 5 3(x - 1)\(^2\) + 3(y + \(\frac{3}{2}\))\(^2\) = \(\frac{49}{4}\) Dividing both sides by 3, we get: (x - 1)\(^2\) + (y + \(\frac{3}{2}\))\(^2\) = (\(\frac{7}{2}\))\(^2\) Comparing this equation to the standard form: (x - h)\(^2\) + (y - k)\(^2\) = r\(^2\) We can see that the center of the circle is at (1, -\(\frac{3}{2}\)) and the radius is (\(\frac{7}{2}\)). Therefore, the answer is (3). (1, -\(\frac{3}{2}\))