Question

Скласти рівняння сфери з центром у точці (5.-9.-12) яка дотикается до осі ординат

Answer 1

(x-5)²+(y+9)²+(z+12)²=R²
R²=x²+z²=5²+(-12)²=25+144=169 ⇒R=√169=13
(x-5)²+(y+9)²+(z+12)²=13²

Answer 2

Equation of a Sphere with Center and Tangent to the y-axis

To find the equation of a sphere with a center at the point (5, -9, -12) that is tangent to the y-axis, we can use the standard equation of a sphere:

Equation of a Sphere: (x - h)^2 + (y - k)^2 + (z - l)^2 = r^2

Where (h, k, l) represents the center of the sphere, and r represents the radius.

In this case, the center of the sphere is given as (5, -9, -12). Since the sphere is tangent to the y-axis, the x-coordinate of the center (h) will be equal to the radius (r).

Therefore, the equation of the sphere can be written as:

(x - 5)^2 + (y - (-9))^2 + (z - (-12))^2 = 5^2

Simplifying the equation further, we get:

(x - 5)^2 + (y + 9)^2 + (z + 12)^2 = 25

So, the equation of the sphere with a center at (5, -9, -12) that is tangent to the y-axis is:

(x - 5)^2 + (y + 9)^2 + (z + 12)^2 = 25