Question

Find the point on the sphere left parenthesis x minus 4 right parenthesis squared(x−4)2 plus+y squared 2 plus+left parenthesis z plus 6 right parenthesis squared(z+6)2 equals=44 nearest to a. the​ xy-plane. b. the point left parenthesis 7 comma 0 comma negative 6 right parenthesis(7,0,−6).

Answer 1

a) It is the set of point in the the circumference with equation
`(x-4)^2+y^2=8`.

b) (10.6, 0, -6 )

Explanation:

a) The centre of the sphere is (4,0,-6) and the radio of the sphere is
`√(44) \sim 6.6`. Since |-6|=6 < 6.6, then the sphere intersect the xy-plane and the intersection is a circumference.

Let's find the equation of the circumference.

The equation of the xy-plane is z=0. Replacing this in the equation of the sphere we have:

`(x-4)^2+y^2+6^2=44`, then
`(x-4)^2+y^2=8`.

b) Observe that the point (7,0,-6) has the same y and z coordinates as the centre and the x coordinate of the point is greater than that of the x coordinate of the centre. Then the point of the sphere nearest to the given point will be at a distance of one radius from the centre, in the positive x direction.

(4+
`√(44)`, 0, -6)= (10.6, 0, -6 )