Find the coordinates of the centre of the circle 3x\(^2\) + 3y\(^2\) - 6x + 9y - 5 = 0
Answer Details
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, -\(\frac{3}{2}\))