Question

If the radius of the sphere is unity ,find the value of d​

Answer 1

Concept Used :-

If the equation of sphere is

$\sf \: {x}^{2} + {y}^{2} + {z}^{2} + 2ux + 2vy + 2wz + d = 0$

then,

Centre of sphere is given by

$\tt \: C = (\dfrac{ - coeff. \: of \: x}{2} ,\dfrac{ - coeff. \: of \: y}{2} , \dfrac{ - coeff. \: of \: z}{2})$

i.e.

$\rm :\implies\:C = ( - u , - v, - w)$

and

Radius of sphere is given by

$\rm :\implies\:r \: = \: \sqrt{ {u}^{2} + {v}^{2} + {w}^{2} - d}$

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$\large\underline\purple{\bold{Solution :- }}$

The equation of sphere is

$\sf \: {2x}^{2} + {2y}^{2} + {2z}^{2} + 3x + 4y + 5z + d = 0$

To represent this equation in general form,

• Divide the whole equation by 2, we get

$\tt \: {x}^{2} + {y}^{2} + {z}^{2} + \dfrac{3}{2} x +2y + \dfrac{5}{2} z + \dfrac{d}{2} = 0$

So,

• Centre of sphere is given by

$\rm :\implies\:C = ( - \dfrac{3}{4} , - 1, - \dfrac{5}{4} )$

and

• radius of sphere is given by

$\rm :\implies\:r \: = \sqrt{ {(\dfrac{ - 3}{4} )}^{2} + {( - 1)}^{2} + {(\dfrac{ - 5}{4} )}^{2} - \dfrac{d}{2} }$

$\rm :\implies\:r \: = \sqrt{\dfrac{9}{16} + 1 + \dfrac{25}{16} - \dfrac{d}{2} }$

$\rm :\implies\:r \: = \: \sqrt{\dfrac{9 + 16 + 25 - 8d}{16} }$

$\rm :\implies\:r \: = \: \sqrt{\dfrac{50 - 8d}{16} }$

$\bullet{ \green{ \bf \: According \: to \: statement,}}$

$\bigstar \: \: { \boxed{ \red{ \tt \:radius \: (r) \: of \: sphere \: = 1 \: unit }}}$

$\rm :\implies\:1 \: = \: \sqrt{\dfrac{50 - 8d}{16} }$

Squaring both sides, we get

$\rm :\implies\:1 = \dfrac{50 - 8d}{16}$

$\rm :\implies\:50 - 8d = 16$

$\rm :\implies\:8d = 34$

$\bigstar \: \: { \boxed{ \red{ \rm :\implies\:d \: = \: \dfrac{17}{4} }}}$

Answer 2

hope you understand

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(prompto)