Question

Find the center and radius of the circle [3x^{2} +3y^{2} -12x+6y-24=0?

Answer 1

The center of the circle given by the equation 3x^2 +3y^2 -12x+6y-24=0 is (2, -1), and the radius is approximately 3.61.

Step-by-step explanation:

To find the center and radius of the circle given by the equation 3x2 +3y2 -12x+6y-24=0, we first need to complete the square for both the x and y terms.

1. Divide the entire equation by 3 to simplify: x2 + y2 - 4x + 2y - 8 = 0.
2. Rearrange the terms to group x's and y's: (x2 - 4x) + (y2 + 2y) = 8.
3. Add the square of half the coefficient of x to both sides to complete the square for x: (x2 - 4x + 4) + (y2 + 2y) = 8 + 4.
4. Add the square of half the coefficient of y to both sides to complete the square for y: (x2 - 4x + 4) + (y2 + 2y + 1) = 8 + 4 + 1.
5. Fully completed squares: (x - 2)2 + (y + 1)2 = 13.
6. The center of the circle is at (2, -1).
7. The radius of the circle is the square root of 13, which is approximately 3.61.