Maths-
General
Easy

Question

Equation of sphere passing through the points (2,0,0),(0,2,0),(0,0,2) and having least possible radius is

  1. left parenthesis x squared plus y squared plus z squared right parenthesis minus 4 left parenthesis x plus y plus z right parenthesis minus 3 equals 0    
  2. 3 open parentheses x to the power of 2 end exponent plus y to the power of 2 end exponent plus z to the power of 2 end exponent close parentheses minus 4 left parenthesis x plus y plus z right parenthesis minus 4 equals 0    
  3. 4 open parentheses x to the power of 2 end exponent plus y to the power of 2 end exponent plus z to the power of 2 end exponent close parentheses minus 3 left parenthesis x plus y plus z right parenthesis minus 5 equals 0    
  4. open parentheses x to the power of 2 end exponent plus y to the power of 2 end exponent plus z to the power of 2 end exponent close parentheses minus left parenthesis x plus y plus z right parenthesis minus 2 equals 0    

hintHint:

Let equation of the sphere be given by x squared space plus space y squared space plus space z squared space plus space 2 u x space plus space 2 v y space plus space 2 w z space plus space d space equals space 0
Find the value of d.

The correct answer is: 3 open parentheses x to the power of 2 end exponent plus y to the power of 2 end exponent plus z to the power of 2 end exponent close parentheses minus 4 left parenthesis x plus y plus z right parenthesis minus 4 equals 0


     Let equation of the sphere be given by x squared space plus space y squared space plus space z squared space plus space 2 u x space plus space 2 v y space plus space 2 w z space plus space d space equals space 0 ...... (1)
    As the sphere passes through points (2, 0, 0), (0, 2, 0) and (0, 0, 2)
    So, we have 
    2 + 4u + d = 0, 
    2 + 4v + d = 0  and
    2 + 4w + d = 0 
    On solving,
    u = v = w = negative 1 fourth left parenthesis d space plus 2 right parenthesis
    If r is the radius of the sphere, then
    r space equals space √ left parenthesis u squared plus space v squared plus space w squared space minus space d right parenthesis

rightwards double arrow r squared space equals space u squared plus space v squared plus space w squared space minus space d

rightwards double arrow r squared space equals 3 over 4 space open parentheses d plus 2 close parentheses squared space minus space d space equals space m left parenthesis l e t right parenthesis
F o r space v a l u e space space r space t o space b e space m i n i m u m
fraction numerator d m over denominator d d end fraction space equals space 0

rightwards double arrow 3 over 4 space 2 open parentheses d plus 2 close parentheses space minus space 1 space equals space 0

rightwards double arrow space d space equals space minus 4 over 3
    Also,

    fraction numerator d squared m over denominator d d squared end fraction space equals space 3 over 2 space equals space p o s i t i v e space a t space d space equals space minus 4 over 3

H e n c e comma space m space i s space m i n i m u m space a t space d space equals space minus 4 over 3

S o comma space s u b s t i t u t i n g space v a l u e space o f space d comma space w e space h a v e space u space equals space v space equals space w space equals space minus 4 over 3

T h e r e f o r e space e q u a t i o n space o f space t h e space s p h e r e space
space x squared space plus space y squared space plus space z squared space minus 4 over 3 left parenthesis x space plus space y space plus z space right parenthesis space minus space 4 over 3 space equals space 0

    Taking LCM
    3 open parentheses x to the power of 2 end exponent plus y to the power of 2 end exponent plus z to the power of 2 end exponent close parentheses minus 4 left parenthesis x plus y plus z right parenthesis minus 4 equals 0
    Thus, equation of sphere passing through the points (2,0,0),(0,2,0),(0,0,2) and having least possible radius is 3 open parentheses x to the power of 2 end exponent plus y to the power of 2 end exponent plus z to the power of 2 end exponent close parentheses minus 4 left parenthesis x plus y plus z right parenthesis minus 4 equals 0
     
     

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