find the standard equation of the sphere that satisfies the stated conditions.

Question

Center (7, 1, 1); radius = 4.(b) Center (1, 0, −1); diameter = 8. (c) Center (−1, 3, 2) and passing through the origin. (d) A diameter has endpoints (−1, 2, 1) and (0, 2, 3)

Answer 1

(a) (x-7)²+(y-1)²+(z-1)²=16

(b) (x-1)²+y²+(z+1)²=(8/2)²=16

(c) (x+1)²+(y-3)²+(z-2)²=r²; sphere passes through origin so (0,0,0) lies on the surface:

1²+(-3)²+(-2)²=1+9+4=14; (x+1)²+(y-3)²+(z-2)²=14

(d) The midpoint of the diameter is the centre of the sphere. The midpoint is the average of the coords:

(-½,2,2) is the centre. Equation of the sphere:

(x+½)²+(y-2)²+(z-2)²=r². The sphere passes through the endpoints of the diameter:

(-1+½)²+(2-2)²+(1-2)²=½²+(2-2)²+(3-2)²=1¼=r². Equation of the sphere:

(x+½)²+(y-2)²+(z-2)²=5/4.