Answer to Question #119196 in Analytic Geometry for Swarohi

Question

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)?

Answer 1

$$(x-x_c)^2+(y-y_c)^2+(z-z_c)^2=R^2$$

$(1,0,0): (1-x_c)^2+(0-y_c)^2+(0-z_c)^2=R^2$

$(0,1,0): (0-x_c)^2+(1-y_c)^2+(0-z_c)^2=R^2$

$(0,0,1): (0-x_c)^2+(0-y_c)^2+(1-z_c)^2=R^2$

$$x_c^2-2x_c+1+y_c^2+z_c^2=R^2$$ $$x_c^2+y_c^2-2y_c+1+z_c^2=R^2$$ $$x_c^2+y_c^2+z_c^2-2z_c+1=R^2$$

$$x_c=y_c=z_c$$ $$3x_c^2-2x_c+1=R^2$$

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

$$({1\over \sqrt{3}}-x_c)^2+({1\over \sqrt{3}}-y_c)^2+({1\over \sqrt{3}}-z_c)^2=R^2$$

Then

$$3({1\over \sqrt{3}}-x_c)^2=3x_c^2-2x_c+1$$

$$1-2\sqrt{3}x_c+3x_c^2=3x_c^2-2x_c+1$$

$$x_c={1\over \sqrt{3}}={\sqrt{3}\over 3}=y_c=z_c$$

$$R^2=3({1\over \sqrt{3}})^2-2({1\over \sqrt{3}})+1=2-{2\over \sqrt{3}}$$

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