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koban [17]
3 years ago
10

Find the equation of the sphere if one of its diameters has endpoints (-8, -3, -10) and (-6, 1, -4) which has been normalized so

that the coefficient of x2 is 1.
Mathematics
1 answer:
alex41 [277]3 years ago
4 0

Answer:

The  equation is  x^2 + y^2    +z^2 + 14 x  + 3y \ + 14z  + 86.25=0

Step-by-step explanation:

From the question we are told that  

    The  diameter endpoints is   (-8, -3, -10) and (-6, 1, -4)

Generally the equation of a sphere with center coordinates (a, b , c ) and  radius  r   is mathematically represented as

           (x -  a )^2  + (y -b )^2  + (z -c)^2 =  r^2

Now since we are given the endpoints of the diameter then we can obtain the center coordinates as follows

         (a, b , c) =  [ \frac{ -8 +(-6)}{2}  ,  \frac{-3 + (1)}{ 2}  , \frac{ -10 + (-4)}{2} ]

         (a, b , c) =  [ -7 ,  -1.5  , -7 ]

Now the length of the diameter is evaluated as

    |d| =  \sqrt{ (-8 - (-6 ))^2 + ( -3 - (1) )^2 + ( -10 - (-4))^2 }

   |d| =  \sqrt{56 }

   |d| =   \sqrt{4 *  14 }

   |d| = 2  \sqrt{  14 }

Now the radius is mathematically represented as

    r = \frac{|d|}{2}

    r = \frac{ 2 \sqrt{14} }{2}

    r =  \sqrt{14}

So

   (x -  -7 )^2  + (y --1.5 )^2  + (z --7)^2 = ( \sqrt{14} )^2

   (x +7 )^2  + (y +1.5 )^2  + (z +7)^2 = 14

    x^2 + 14 x + 49  + y^2 + 3y + 2.25   +z^2 14z + 49 = 14

    x^2 + y^2    +z^2 + 14 x  + 3y \ + 14z  + 86.25=0

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