Question

. In each part, find the standard equation of the sphere that
satisfies the stated conditions.
(a) 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

Step-by-step explanation:

standard equation of sphere    , (x-a)²+ (y-b)²+( z-c)² = r² where a.b.c are the centre of the sphere r is radius and x,y,z are the coordintes of the point on the surface of the sphere

(x-7)²+( y-1)²+(z-1)²=16

x²+49-14x+y²+1-2y+z²+1-2z=16

x²+y²+z²-14x-2y-2z+35=0

(b) same as above diameter = 8 ,radius = 4 and centre is given

(c) (x-1)²+(y-3)²+(z-2)² = r²  it passes through (0,0,0)

     (0-1)²+( 0-3)²+(0-2)²=r²  ⇒ r =√14   now we have centre and radius same as above

(d) centre is mid point of the end points of diameter

(1+0/2)=x   ,(2+2/2) =y   (1+3/2) =z ⇒   (x,y,z) = (1/2,2,2)  and

radius = distance between centre and any end points of the diameter

so radius =  √(1/2-1)²+(2-2)²+(2-1)²   =   √5/2 now we can find the equation of the sphere