Final answer:
The intersection of a sphere centered at (h, k, l) with a radius r and a plane described by the equation Ax + By + Cz + D = 0 is given by substituting the plane's equation into the sphere's equation. If the z-value is constant, the intersection is a circle represented by (x - h)² + (y - k)² = r² - (constant - l)², where the circle lies in the yz-plane.
Step-by-step explanation:
The student's question asks for an equation that describes the intersection of a sphere with a plane. The intersection of a sphere and a plane is a circle if the plane cuts through the sphere or a point if the plane is tangent to the sphere. To provide an equation for this intersection, suppose we have a sphere with center (h, k, l) and radius r, described by the equation (x - h)² + (y - k)² + (z - l)² = r². The equation of a plane can be written as Ax + By + Cz + D = 0, where A, B, and C are the coefficients that define the plane's orientation, and D is the distance along the normal from the origin to the plane.
Let's assume we know the coefficients A, B, C, and D for our specific plane. The intersection is discovered by substituting the plane's equation into the sphere's equation and solving for a circle. This results in the equation of a circle after restricting z to be a constant, the value which satisfies the plane equation when x and y are variables.
Thus, if the plane intersects the sphere such that z = constant, we will get a circle in the y-z plane with the equation:
(x - h)² + (y - k)² = r² - (constant - l)²
This equation represents the projection of the sphere onto the yz-plane, showing the locus of points that form the circle, which is the intersection of the sphere with the specific plane.