Question #119196
Find the centre and radius of the sphere passing through (1,0,0),(0,1,0),(0,0,1) and (1/√3,1/√3,1/√3)?
1
Expert's answer
2020-06-01T18:10:27-0400
(xxc)2+(yyc)2+(zzc)2=R2(x-x_c)^2+(y-y_c)^2+(z-z_c)^2=R^2

(1,0,0):(1xc)2+(0yc)2+(0zc)2=R2(1,0,0): (1-x_c)^2+(0-y_c)^2+(0-z_c)^2=R^2

(0,1,0):(0xc)2+(1yc)2+(0zc)2=R2(0,1,0): (0-x_c)^2+(1-y_c)^2+(0-z_c)^2=R^2

(0,0,1):(0xc)2+(0yc)2+(1zc)2=R2(0,0,1): (0-x_c)^2+(0-y_c)^2+(1-z_c)^2=R^2



xc22xc+1+yc2+zc2=R2x_c^2-2x_c+1+y_c^2+z_c^2=R^2xc2+yc22yc+1+zc2=R2x_c^2+y_c^2-2y_c+1+z_c^2=R^2xc2+yc2+zc22zc+1=R2x_c^2+y_c^2+z_c^2-2z_c+1=R^2


xc=yc=zcx_c=y_c=z_c3xc22xc+1=R23x_c^2-2x_c+1=R^2

(1/3,1/3,1/3):(1/\sqrt{3}, 1/\sqrt{3}, 1/\sqrt{3}):


(13xc)2+(13yc)2+(13zc)2=R2({1\over \sqrt{3}}-x_c)^2+({1\over \sqrt{3}}-y_c)^2+({1\over \sqrt{3}}-z_c)^2=R^2

Then


3(13xc)2=3xc22xc+13({1\over \sqrt{3}}-x_c)^2=3x_c^2-2x_c+1

123xc+3xc2=3xc22xc+11-2\sqrt{3}x_c+3x_c^2=3x_c^2-2x_c+1

xc=13=33=yc=zcx_c={1\over \sqrt{3}}={\sqrt{3}\over 3}=y_c=z_c

R2=3(13)22(13)+1=223R^2=3({1\over \sqrt{3}})^2-2({1\over \sqrt{3}})+1=2-{2\over \sqrt{3}}

C(33,33,33),R=223C({\sqrt{3}\over 3}, {\sqrt{3}\over 3}, {\sqrt{3}\over 3}), R=\sqrt{2-{2\over \sqrt{3}}}


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Comments

Assignment Expert
02.06.20, 21:42

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Swarohi
02.06.20, 05:19

Thank you so much

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