Final answer:
To find the equation of a sphere that passes through the origin with a center at (1, 2, 3), calculate the radius using the distance formula to get √14. Plug in the center coordinates and radius into the standard sphere equation to get (x - 1)² + (y - 2)² + (z - 3)² = 14.
Step-by-step explanation:
The question asks us to find the equation of a sphere that passes through the origin and has its center at the point (1, 2, 3). A sphere's equation is given by (x - h)² + (y - k)² + (z - l)² = r², where (h, k, l) is the center of the sphere and r is the radius. Since the sphere passes through the origin, we can use this point to determine the radius.
First, calculate the radius by finding the distance from the center to the origin using the distance formula: r = √((1-0)² + (2-0)² + (3-0)²). Thus, r = √(1+4+9) = √14. Next, plug the center (h, k, l) = (1, 2, 3) and r = √14 into the sphere's equation: (x - 1)² + (y - 2)² + (z - 3)² = (√14)², which simplifies to (x - 1)² + (y - 2)² + (z - 3)² = 14. This is the equation of the required sphere.
To find the equation of the sphere, we can use the equation for the distance between a point (x, y, z) and the center of the sphere (a, b, c) as follows:
d = sqrt((x-a)^2 + (y-b)^2 + (z-c)^2)
Since the sphere passes through the origin, the point (0, 0, 0) is on the sphere. Plugging in these values, we get:
d = sqrt((0-1)^2 + (0-2)^2 + (0-3)^2)
d = sqrt((-1)^2 + (-2)^2 + (-3)^2)
d = sqrt(14)
So the equation of the sphere is x^2 + y^2 + z^2 = 14.