Question

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.

Answer 1

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

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

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

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

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