Questions: Write the equation of the sphere in standard form. x^2+y^2+z^2+6x-2y-4z=11 Find its center and radius. center (x, y, z)=() radius

Solution Steps

To write the equation of the sphere in standard form, we need to complete the square for each of the variables \(x\), \(y\), and \(z\). This will allow us to express the equation in the form \((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\), where \((h, k, l)\) is the center of the sphere and \(r\) is the radius.

The given equation of the sphere is:

\[ x^{2} + y^{2} + z^{2} + 6x - 2y - 4z = 11 \]

To write the equation in standard form, we need to complete the square for each variable. The standard form of a sphere's equation is:

\[ (x - h)^2 + (y - k)^2 + (z - l)^2 = r^2 \]

where \((h, k, l)\) is the center and \(r\) is the radius.

The terms involving \(x\) are \(x^2 + 6x\).

1. Take the coefficient of \(x\), which is 6, divide by 2, and square it: \(\left(\frac{6}{2}\right)^2 = 9\).
2. Add and subtract 9 inside the equation:

\[ x^2 + 6x = (x^2 + 6x + 9) - 9 = (x + 3)^2 - 9 \]

The terms involving \(y\) are \(y^2 - 2y\).

1. Take the coefficient of \(y\), which is -2, divide by 2, and square it: \(\left(\frac{-2}{2}\right)^2 = 1\).
2. Add and subtract 1 inside the equation:

\[ y^2 - 2y = (y^2 - 2y + 1) - 1 = (y - 1)^2 - 1 \]

The terms involving \(z\) are \(z^2 - 4z\).

1. Take the coefficient of \(z\), which is -4, divide by 2, and square it: \(\left(\frac{-4}{2}\right)^2 = 4\).
2. Add and subtract 4 inside the equation:

\[ z^2 - 4z = (z^2 - 4z + 4) - 4 = (z - 2)^2 - 4 \]

Substitute the completed squares back into the original equation:

\[ (x + 3)^2 - 9 + (y - 1)^2 - 1 + (z - 2)^2 - 4 = 11 \]

Combine the constants on the right side:

\[ (x + 3)^2 + (y - 1)^2 + (z - 2)^2 = 11 + 9 + 1 + 4 \]

\[ (x + 3)^2 + (y - 1)^2 + (z - 2)^2 = 25 \]

The equation is now in standard form:

\[ (x + 3)^2 + (y - 1)^2 + (z - 2)^2 = 25 \]

From this, we can identify:

• The center \((h, k, l)\) is \((-3, 1, 2)\).
• The radius \(r\) is \(\sqrt{25} = 5\).

Final Answer