Complete the square to find the center and the radius of the circle.
3x^2+3y^2-12x+24y+15=0
Connor's question from before
http://www.jiskha.com/display.cgi?id=1283297479
notice my correction at the end
To complete the square for the given equation, follow these steps:
Step 1: Group the terms with x and y together:
3x^2-12x + 3y^2+24y = -15
Step 2: Complete the square for the x-terms:
To complete the square for the x-terms, take half of the coefficient of x (which is -12 in this case), square it, and add it to both sides of the equation:
3x^2-12x + (-12/2)^2 = -15 + (-12/2)^2
3x^2-12x + 36 = -15 + 36
Step 3: Complete the square for the y-terms:
To complete the square for the y-terms, take half of the coefficient of y (which is 24 in this case), square it, and add it to both sides of the equation:
3y^2+24y + (24/2)^2 = -15 + 36 + (24/2)^2
3y^2+24y + 144 = -15 + 36 + 144
Step 4: Simplify the constants on the right side:
3x^2-12x + 36 = 21
3y^2+24y + 144 = 165
Step 5: Rewrite the left side as a squared binomial:
(x - h)^2 = 21
(y - k)^2 = 165
Step 6: Write the equation in standard form:
(x - h)^2 + (y - k)^2 = r^2
Comparing the equations, we have:
(x - h)^2 + (y - k)^2 = 21
(x - h)^2 + (y - k)^2 = 165
The two equations are in the standard form of a circle, where the center of the circle is given by the point (h, k) and the radius is given by r.
Therefore, the center and radius of the circle are as follows:
Center: (h, k)
Radius: √r
By comparing the equations, we can see that the center and radius are not given explicitly, so we cannot determine them from the given equation 3x^2 + 3y^2 - 12x + 24y + 15 = 0.