Problem 13 - Algebra and Trigonometry

Question

\(11-14\) . Find an equation of a sphere with the given radius \(r\) and center \(C .\) $$ r=\sqrt{6} ; \quad Q(3,-1,0) $$

Step-by-Step Solution

Answer

The equation of the sphere is \((x-3)^2 + (y+1)^2 + z^2 = 6\).

Step 1: Write the General Equation of a Sphere

The general equation of a sphere in the coordinate space is \((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\). Here, \((h, k, l)\) is the center of the sphere, and \(r\) is the radius.

Step 2: Identify the Center Coordinates

From the problem, we have the center \(Q(3, -1, 0)\). Therefore, \(h = 3\), \(k = -1\), and \(l = 0\).

Step 3: Compute the Radius Squared

The radius given is \(r = \sqrt{6}\). To find \(r^2\), we compute: \((\sqrt{6})^2 = 6\).

Step 4: Substitute into the Equation of the Sphere

Substitute \(h = 3\), \(k = -1\), \(l = 0\), and \(r^2 = 6\) into the general equation: \((x-3)^2 + (y+1)^2 + (z-0)^2 = 6\). This simplifies to the equation of the sphere.

Key Concepts

Coordinate Geometry

Coordinate geometry helps describe geometric figures using a coordinate system. Here, each point in space is represented using three coordinates, which describe the position along the x, y, and z axes. In the context of a sphere, the coordinates of the center and the radius allow us to express the sphere's properties and location in a three-dimensional space.

Let's analyze how this applies to the equation of a sphere. The sphere's equation typically appears as \((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\). This equation connects the geometric properties of the sphere to its coordinate representation:

• \(x, y, z\) are the coordinates of any generic point on the surface of the sphere.
• \(h, k, l\) are the center coordinates of the sphere.
• \(r\) is the radius of the sphere.

Understanding coordinate geometry is crucial for analyzing and solving problems related to the position and dimensions of spherical objects in space. By representing them through equations, it's easier to handle calculations and determine relationships with other geometric figures.

Radius of a Sphere

The radius of a sphere is the distance from the center of the sphere to any point on its surface. In a three-dimensional coordinate system, this radius is an essential part because it defines the size of the sphere. The radius is constant for all points on the surface of the sphere, ensuring our equation remains valid across the entire surface.

In our exercise, we see the radius given as \(r = \sqrt{6}\). To use this in the equation of the sphere, we substitute the square of the radius, \(r^2\), instead. We calculate:
- For our example, \((\sqrt{6})^2 = 6\).
By understanding this step, we can see how the radius affects the overall size of the sphere when expressed in the coordinate geometry format. The square of the radius is what balances the equation and ensures it remains true for every point \((x, y, z)\) that lies on the sphere's surface.

Center of a Sphere

The center of a sphere in coordinate geometry is a fixed point defined by its coordinates, typically denoted as \((h, k, l)\). It represents the mid-point from which the radius extends equally in all directions. Understanding the center is key as it situates the sphere within the coordinate geometry framework.

For instance, in our problem, the center of the sphere is given by \(Q(3, -1, 0)\). These coordinates are plugged directly into the sphere equation:

• \(h = 3\)
• \(k = -1\)
• \(l = 0\)

As we substitute these into \((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\), we naturally derive the equation of our specific sphere. The center point \((h, k, l)\) is critical because it not only determines the position but also affects how the sphere interacts with other geometrical entities, such as planes and other spheres.