Find the point on the sphere (x+5)^2 + y^2 + (z−9)^2 = 99 nearest to (a) the xy-plane. (b) the point (−9,0,9).
Accepted Solution
A:
Answer:a) Since the sphere intersects the xy-plane then the set of points of the sphere nearest to the xy-plane is the set of points in the circumference [tex](x+5)^2+y^2=18[/tex].b)(-14.9, 0, 9 )Step-by-step explanation:a) The centre of the sphere is (-5,0,-9) and the radio of the sphere is [tex]\sqrt{99} \sim 9.9[/tex]. Since |-9|=9 < 9.9, 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:[tex](x+5)^2+y^2+9^2=99[/tex], then [tex](x+5)^2+y^2=18[/tex].b) Observe that the point (-9,0,9) has the same y and z coordinates as the centre and the x coordinate of the point is smaller 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 negative x direction. (-5-[tex]\sqrt{99}[/tex], 0, 9)= (-14.9, 0, 9 )