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 equation of the sphere is \((x-1)^2 + (y-1)^2 + (z-4)^2 = 50\).

Step 1: Understand the Problem

To find the equation of a sphere, we use the general formula \((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\), where \((h, k, l)\) is the center and \(r\) is the radius. We know the center of the sphere is \((1,1,4)\) and it is tangent to the plane \(x+y=12\).

Step 2: Determine the Radius

The radius of the sphere equals the perpendicular distance from the center to the plane, as the sphere is tangent to the plane. The formula for the perpendicular distance \(d\) from a point \((x_1, y_1, z_1)\) to a plane \(Ax + By + Cz + D = 0\) is \[ d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}} \].

Step 3: Apply the Distance Formula

Using the point \((1, 1, 4)\) and the plane equation \(x + y - 12 = 0\) which can be rewritten as \(x + y + 0z - 12 = 0\), we calculate the distance: \[ d = \frac{|1\cdot1 + 1\cdot1 + 0\cdot4 - 12|}{\sqrt{1^2 + 1^2 + 0^2}} = \frac{|1 + 1 - 12|}{\sqrt{2}} = \frac{10}{\sqrt{2}} = 5\sqrt{2}. \]Thus, the radius \(r\) is \(5\sqrt{2}\).

Step 4: Write the Equation of the Sphere

Substitute the center \((1, 1, 4)\) and the radius \(r = 5\sqrt{2}\) into the sphere equation: \[ (x - 1)^2 + (y - 1)^2 + (z - 4)^2 = (5\sqrt{2})^2 \].This simplifies to: \( (x - 1)^2 + (y - 1)^2 + (z - 4)^2 = 50 \).

Key Concepts

Tangent Plane

A tangent plane is a flat surface that just "touches" or intersects another surface at a single point. In the context of a sphere, particularly our example where the sphere is tangent to a plane, the tangent plane touches the sphere at exactly one point. This point is where the plane and the sphere are closest, hence the name "tangent."
For a plane to be tangent to a sphere, the perpendicular distance from the sphere's center to the plane must be equal to the sphere's radius. This ensures the sphere "grazes" the plane without cutting through it. In our example, this perpendicular distance from the center \(1, 1, 4\) to the plane \(x + y = 12\) provides the radius of the sphere.

Center of the Sphere

The center of a sphere is simply the point from which every point on the surface is equidistant. Think of it like the nucleus of an atom, evenly balancing the surrounding electrons, or in this case, the spherical surface.
For our problem, the center is given as \(1, 1, 4\). This means that every point on the sphere's surface is equidistant from this center point. The center acts as a reference point not only for detailing the location of the sphere but also for determining other features like the radius.
Since the center is positioned at \(1, 1, 4\), each point on the plane \(x+y=12\), at its closest, should be exactly one radius away from this center.

Radius of a Sphere

The radius of a sphere is the fixed distance from its center to any point on its surface. It acts as a key parameter in defining the geometry of a sphere.

• To find the radius given a point \(x_1, y_1, z_1\) and a tangent plane \(Ax + By + Cz + D = 0\), we calculate the perpendicular distance from the point to the plane.
• In our solution, this is determined using the formula: \[ d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}\]

Using the example of the center \(1, 1, 4\) and the plane's equation rewritten as \(x + y + 0z - 12 = 0\), the radius comes out to be \(5\sqrt{2}\), representing the distance from the center to this tangent plane.

Perpendicular Distance to a Plane

Determining the perpendicular distance from a point to a plane is crucial in many geometry problems, especially when dealing with spheres and planes. This distance gives powerful insights, particularly about how shapes like spheres interact with planes.
The perpendicular distance formula \[ d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}} \] helps find the shortest distance between a point \(x_1, y_1, z_1\) and the plane \(Ax + By + Cz + D = 0\). In scenario-based problems such as our exercise, where the sphere is tangent to a plane, this distance is essentially the radius of the sphere.
This instance clearly outlines why the perpendicular distance is vital, as it drives the relationship between the sphere and its tangent plane.