Show that the equation represent a sphere, and find it center and radius 3x^2 +3y^2 + 3z^2=10+6y+12z
Remember what you did with circles in Algebra II -- complete the squares.
3x^2 +3y^2 + 3z^2=10+6y+12z
3x^2 + 3y^2-6y + 3z^2-12z = 10
3x^2 + 3(y^2-2y+1) + 3(z^2-4z+4) = 10 + 3*1 + 3*4
and so on to the end.
To determine whether the equation represents a sphere, we need to convert it to the standard form of a sphere equation, which is:
(x - a)^2 + (y - b)^2 + (z - c)^2 = r^2,
where (a, b, c) represents the center of the sphere and r represents the radius.
Let's begin by rearranging the given equation:
3x^2 + 3y^2 + 3z^2 = 10 + 6y + 12z
Dividing the entire equation by 3, we get:
x^2 + y^2 + z^2 - 2y - 4z = 10/3
To complete the square, we will focus on the x, y, and z terms separately.
For the x-term:
x^2 = x^2 + 0x + 0^2 = (x - 0)^2
For the y-term:
y^2 - 2y = (y^2 - 2y + 1) - 1 = (y - 1)^2 - 1
For the z-term:
z^2 - 4z = (z^2 - 4z + 4) - 4 = (z - 2)^2 - 4
Substituting these values back into the equation, we have:
(x - 0)^2 + [(y - 1)^2 - 1] + [(z - 2)^2 - 4] = 10/3
Simplifying further:
x^2 + (y - 1)^2 + (z - 2)^2 - 1 - 4 = 10/3
Combining constants:
x^2 + (y - 1)^2 + (z - 2)^2 - 5 = 10/3
Now, we can compare this equation to the standard form, where (a, b, c) represents the center and r^2 is the square of the radius. From the equation, we can deduce that the sphere's center is at (0, 1, 2) and the radius is √(10/3 + 5).
Therefore, the equation represents a sphere with center (0, 1, 2) and radius √(10/3 + 5).
To show that the given equation represents a sphere, we need to rearrange it into the standard form of a sphere equation, which is:
(x - h)^2 + (y - k)^2 + (z - l)^2 = r^2
where (h, k, l) represents the center of the sphere, and r represents the radius.
Let's rearrange the given equation step by step:
3x^2 + 3y^2 + 3z^2 = 10 + 6y + 12z
First, we want to group the x, y, and z terms separately:
3x^2 - 6y + 3y^2 + 3z^2 - 12z = 10
Next, we complete the square individually for x, y, and z terms.
For the x term:
Factor out the coefficient of x^2, which is 3:
3(x^2 - 2x) + 3y^2 + 3z^2 - 12z = 10
To complete the square for x, we take half of the coefficient of x, square it, and add it inside the parenthesis. So half of -2 is -1, and (-1)^2 is 1. Add 1 inside the parenthesis and subtract 3 times the added value to keep the equation balanced.
3(x^2 - 2x + 1) + 3y^2 + 3z^2 - 12z - 3(1) = 10 - 3
This simplifies to:
3(x - 1)^2 + 3y^2 + 3z^2 - 12z - 3 = 7
Repeat the same process for y and z terms:
For the y term:
3(x - 1)^2 + 3(y^2 - 2y) + 3z^2 - 12z - 3 = 7
Take half of the coefficient of y, which is -2, square it to get 4, and add it inside the parenthesis. Subtract 3 times the added value to balance the equation.
3(x - 1)^2 + 3(y^2 - 2y + 1) - 3(4) + 3z^2 - 12z - 3 = 7 - 12
This simplifies to:
3(x - 1)^2 + 3(y - 1)^2 + 3z^2 - 12z - 15 = -5
For the z term:
3(x - 1)^2 + 3(y - 1)^2 + 3(z^2 - 4z) - 3(2^2) - 15 = -5
Take half of the coefficient of z, which is -4, square it to get 16, and add it inside the parenthesis. Subtract 3 times the added value to balance the equation.
3(x - 1)^2 + 3(y - 1)^2 + 3(z^2 - 4z + 4) - 3(16) - 15 = -5 - 3(16)
This simplifies to:
3(x - 1)^2 + 3(y - 1)^2 + 3(z - 2)^2 = 0
Now, we can see that the equation is in the standard form of a sphere. The center of the sphere is (1, 1, 2), and the radius is 0 (since the coefficient of (z - 2)^2 is 0).
Therefore, the given equation represents a degenerate sphere that consists only of a single point (1, 1, 2).