Find the radius of the circle 2x\(^2\) - 4x + 2y\(^2\) - 6y -2 = 0.
Answer Details
To find the radius of a circle from its equation in general form, we need to rewrite the equation in the standard form of a circle, which is:
(x - h)\(^2\) + (y - k)\(^2\) = r\(^2\)
Where (h, k) is the center of the circle, and r is its radius.
To do this, we complete the square for both x and y terms in the given equation. We start by rearranging the terms as follows:
2x\(^2\) - 4x + 2y\(^2\) - 6y -2 = 0
2x\(^2\) - 4x + 2y\(^2\) - 6y = 2
Now, we need to add and subtract appropriate constants to complete the square for x and y terms separately. For the x terms, we take half of the coefficient of x (-4/2 = -2) and square it to get 4. So, we add and subtract 4 to the equation:
2x\(^2\) - 4x + 4 - 4 + 2y\(^2\) - 6y = 2
We can now group the first three terms and factor it as a perfect square:
2(x - 1)\(^2\) + 2y\(^2\) - 6y - 2 = 0
For the y terms, we take half of the coefficient of y (-6/2 = -3) and square it to get 9. So, we add and subtract 9 to the equation:
2(x - 1)\(^2\) + 2(y - 3)\(^2\) - 2 - 9 = 0
2(x - 1)\(^2\) + 2(y - 3)\(^2\) = 11
Now, we have the equation in the standard form of a circle, where the center is at (1, 3) and the radius is the square root of 11/2. We can simplify this expression to get:
sqrt(11/2) = sqrt(11)/sqrt(2) = sqrt(2)*sqrt(11)/2
Therefore, the answer is option (C), 17/√2.