Problem 40 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the equation of the sphere with center \((1,1,4)\) that is tangent to the plane \(x+y=12\).

Step-by-Step Solution

Answer

The sphere's equation is \\( (x-1)^2 + (y-1)^2 + (z-4)^2 = 50 \\\).

Step 1: Understand the Sphere Equation

The equation of a sphere with center \( (h, k, l) \) and radius \( r \) is given as \( (x-h)^2 + (y-k)^2 + (z-l)^2 = r^2 \). For our problem, the center is \( (1, 1, 4) \), so the equation becomes \( (x-1)^2 + (y-1)^2 + (z-4)^2 = r^2 \). We need to find the radius \( r \).

Step 2: Analyze Tangency Condition

A sphere is tangent to a plane if the shortest distance from the center of the sphere to the plane equals the radius. The plane given by equation is \( x+y=12 \). We find the distance from the center \( (1,1,4) \) to this plane using the formula: \[ \text{Distance} = \frac{|Ax_1+By_1+Cz_1+D|}{\sqrt{A^2+B^2+C^2}} \]where \( Ax+By+Cz+D = 0 \) is the plane equation. Rewriting the plane equation \( x+y-12=0 \), we get \( A=1, B=1, C=0, D=-12 \).

Step 3: Calculate Distance from Center to Plane

Substitute the center point \( (1,1,4) \) into the distance formula:\[\text{Distance} = \frac{|1*1 + 1*1 + 0*4 - 12|}{\sqrt{1^2+1^2+0^2}} = \frac{|1+1-12|}{\sqrt{2}} = \frac{| -10 |}{\sqrt{2}} = \frac{10}{\sqrt{2}} = 5\sqrt{2}\]This distance, \( 5\sqrt{2} \), is the radius \( r \).

Step 4: Write the Sphere's Equation

Now that we know the radius \( r = 5\sqrt{2} \), we substitute it into the sphere equation. Thus, the sphere's equation is \[(x-1)^2 + (y-1)^2 + (z-4)^2 = (5\sqrt{2})^2\] which simplifies to \[(x-1)^2 + (y-1)^2 + (z-4)^2 = 50\].

Key Concepts

Tangency Condition

When we talk about a sphere being tangent to a plane, we mean that the sphere just touches the plane at one single point. This condition is crucial in many geometric and math problems.
For this tangency to hold true:

• The shortest distance from the center of the sphere to the plane must be exactly equal to the radius of the sphere.
• If this condition is met, then the sphere doesn't penetrate the plane, it just "kisses" it.

Understanding this concept helps solve many problems involving spheres and planes because it sets a specific condition that must be satisfied for tangency.

Distance from Point to Plane

Calculating the distance from a point to a plane is a fundamental concept in geometry. This distance can be found using a specific formula, especially when dealing with planes described by equations like Ax + By + Cz + D = 0.
To calculate this distance, follow these steps:

• Identify the plane equation and the coefficients, A, B, C, and D.
• Substitute the coordinates of the point into the formula: \[\text{Distance} = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}.\]
• This formula gives you the shortest, or perpendicular distance from the point to the plane.

In our problem, using the coordinates of the center of the sphere and the plane equation, we determine this crucial distance, which should equal the radius when tangency is involved.

Center of the Sphere

The center of a sphere is one of its most important properties, often denoted as \((h, k, l)\). It acts as the reference point from which every point on the surface of the sphere is equidistant.
For our sphere problem:

• The center is given as \((1, 1, 4)\).
• This designated center is critical when calculating the sphere's radius, especially in relation to other geometric structures like planes.

Thus, the center serves as a fixed point from which key characteristics of the sphere are analyzed, including its equation, and interactions like tangency with other components.

Radius Calculation

The radius of a sphere determines its size and is half the diameter. When a sphere is tangent to another geometric entity, such as a plane, the radius becomes especially significant because it exactly equals the distance from the sphere's center to that tangent surface.
Ways to determine the radius include:

• Using the standard formula for distance (as we calculated from the center of the sphere to the plane).
• Performing direct calculations when the conditions (like tangency) precisely define this value.

In our exercise, after calculating the distance from the sphere's center to the plane, we identified this distance of \(5\sqrt{2}\) as the exact radius. With this radius, we can derive the complete equation of the sphere accurately.