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Question Number 182369 by mathocean1 last updated on 08/Dec/22

Find radius and center of (S):  (S): { ((x+y+z=4)),((y^2 +yz+z^2 =4(y+z))) :}

$${Find}\:{radius}\:{and}\:{center}\:{of}\:\left({S}\right): \\ $$$$\left({S}\right):\begin{cases}{{x}+{y}+{z}=\mathrm{4}}\\{{y}^{\mathrm{2}} +{yz}+{z}^{\mathrm{2}} =\mathrm{4}\left({y}+{z}\right)}\end{cases} \\ $$

Commented by a.lgnaoui last updated on 09/Dec/22

y+z=4−x                   (1)  (y+z)^2 −yz=4(y+z)  (2)  4−x)^2 −yz=4(4−x)  yz=x^2 −4x    y+z=4−x                    (3)  yz    =x(x−4)              (4)   y=4−x−z  yz=z(4−x−z)=x(x−4)  =(4−x)z−z^2 =x(x−4)  z^2   −(4−x)z−x(x−4)=0     (5)             (4−x)^2 +4x(x−4)=0  16−8x+x^2 +4x^2 −16x=0     5x^2 −24x+16=0  12^2 −80=144−80=64=8^2   x=((12±8)/5)      x=(4,(4/5))  z=  1•    x=4     y+z=0    y=−z            y^2 =0        y=z=0   2•  x=(4/5)   y+z=((20)/5)−(4/5)=((16)/5)      y=((16)/5)−z      yz=x^2 −4x=((4/5))^2 −4((4/5))=((16−80)/(25))  yz=−((64)/(25))   = z(((16)/5)−z)        z^2 −((16)/5)z−((64)/(25))=0   z=(8/5)±((8(√2))/5)  z=(8/5)(1±(√2))  y=((16)/5)±(8/5)(1±(√2))  Conclusion:  A(4,0,0)   B((4/5),(8/5)(1+(√2) ),(8/5)(1−(√2) ) )  C((4/5),(8/5)  (1−(√2) ),(8/5)(1+(√2) ))   (A suivre  )......

$$\mathrm{y}+\mathrm{z}=\mathrm{4}−\mathrm{x}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{1}\right) \\ $$$$\left(\mathrm{y}+\mathrm{z}\right)^{\mathrm{2}} −\mathrm{yz}=\mathrm{4}\left(\mathrm{y}+\mathrm{z}\right)\:\:\left(\mathrm{2}\right) \\ $$$$\left.\mathrm{4}−{x}\right)^{\mathrm{2}} −{y}\mathrm{z}=\mathrm{4}\left(\mathrm{4}−\mathrm{x}\right) \\ $$$$\mathrm{yz}=\mathrm{x}^{\mathrm{2}} −\mathrm{4x} \\ $$$$ \\ $$$$\mathrm{y}+\mathrm{z}=\mathrm{4}−\mathrm{x}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{3}\right) \\ $$$$\mathrm{yz}\:\:\:\:=\mathrm{x}\left(\mathrm{x}−\mathrm{4}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{4}\right) \\ $$$$\:\mathrm{y}=\mathrm{4}−\mathrm{x}−\mathrm{z} \\ $$$$\mathrm{yz}=\mathrm{z}\left(\mathrm{4}−\mathrm{x}−\mathrm{z}\right)=\mathrm{x}\left(\mathrm{x}−\mathrm{4}\right) \\ $$$$=\left(\mathrm{4}−\mathrm{x}\right)\mathrm{z}−\mathrm{z}^{\mathrm{2}} =\mathrm{x}\left(\mathrm{x}−\mathrm{4}\right) \\ $$$$\mathrm{z}^{\mathrm{2}} \:\:−\left(\mathrm{4}−\mathrm{x}\right)\mathrm{z}−\mathrm{x}\left(\mathrm{x}−\mathrm{4}\right)=\mathrm{0}\:\:\:\:\:\left(\mathrm{5}\right)\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\left(\mathrm{4}−\mathrm{x}\right)^{\mathrm{2}} +\mathrm{4x}\left(\mathrm{x}−\mathrm{4}\right)=\mathrm{0} \\ $$$$\mathrm{16}−\mathrm{8x}+\mathrm{x}^{\mathrm{2}} +\mathrm{4x}^{\mathrm{2}} −\mathrm{16x}=\mathrm{0} \\ $$$$\:\:\:\mathrm{5x}^{\mathrm{2}} −\mathrm{24x}+\mathrm{16}=\mathrm{0} \\ $$$$\mathrm{12}^{\mathrm{2}} −\mathrm{80}=\mathrm{144}−\mathrm{80}=\mathrm{64}=\mathrm{8}^{\mathrm{2}} \\ $$$$\mathrm{x}=\frac{\mathrm{12}\pm\mathrm{8}}{\mathrm{5}}\:\:\:\:\:\:\mathrm{x}=\left(\mathrm{4},\frac{\mathrm{4}}{\mathrm{5}}\right) \\ $$$$\mathrm{z}= \\ $$$$\mathrm{1}\bullet\:\:\:\:\mathrm{x}=\mathrm{4}\:\:\:\:\:\mathrm{y}+\mathrm{z}=\mathrm{0}\:\:\:\:\mathrm{y}=−\mathrm{z}\:\:\: \\ $$$$\:\:\:\:\:\:\:\mathrm{y}^{\mathrm{2}} =\mathrm{0}\:\:\:\:\:\:\:\:\mathrm{y}=\mathrm{z}=\mathrm{0} \\ $$$$\:\mathrm{2}\bullet\:\:\mathrm{x}=\frac{\mathrm{4}}{\mathrm{5}}\:\:\:\mathrm{y}+\mathrm{z}=\frac{\mathrm{20}}{\mathrm{5}}−\frac{\mathrm{4}}{\mathrm{5}}=\frac{\mathrm{16}}{\mathrm{5}} \\ $$$$\:\:\:\:\mathrm{y}=\frac{\mathrm{16}}{\mathrm{5}}−\mathrm{z}\:\:\:\: \\ $$$${yz}={x}^{\mathrm{2}} −\mathrm{4}{x}=\left(\frac{\mathrm{4}}{\mathrm{5}}\right)^{\mathrm{2}} −\mathrm{4}\left(\frac{\mathrm{4}}{\mathrm{5}}\right)=\frac{\mathrm{16}−\mathrm{80}}{\mathrm{25}} \\ $$$${yz}=−\frac{\mathrm{64}}{\mathrm{25}}\:\:\:=\:{z}\left(\frac{\mathrm{16}}{\mathrm{5}}−{z}\right) \\ $$$$\:\:\:\:\:\:{z}^{\mathrm{2}} −\frac{\mathrm{16}}{\mathrm{5}}{z}−\frac{\mathrm{64}}{\mathrm{25}}=\mathrm{0}\:\:\:{z}=\frac{\mathrm{8}}{\mathrm{5}}\pm\frac{\mathrm{8}\sqrt{\mathrm{2}}}{\mathrm{5}} \\ $$$${z}=\frac{\mathrm{8}}{\mathrm{5}}\left(\mathrm{1}\pm\sqrt{\mathrm{2}}\right) \\ $$$${y}=\frac{\mathrm{16}}{\mathrm{5}}\pm\frac{\mathrm{8}}{\mathrm{5}}\left(\mathrm{1}\pm\sqrt{\mathrm{2}}\right) \\ $$$${Conclusion}: \\ $$$${A}\left(\mathrm{4},\mathrm{0},\mathrm{0}\right)\:\:\:{B}\left(\frac{\mathrm{4}}{\mathrm{5}},\frac{\mathrm{8}}{\mathrm{5}}\left(\mathrm{1}+\sqrt{\mathrm{2}}\:\right),\frac{\mathrm{8}}{\mathrm{5}}\left(\mathrm{1}−\sqrt{\mathrm{2}}\:\right)\:\right) \\ $$$${C}\left(\frac{\mathrm{4}}{\mathrm{5}},\frac{\mathrm{8}}{\mathrm{5}}\:\:\left(\mathrm{1}−\sqrt{\mathrm{2}}\:\right),\frac{\mathrm{8}}{\mathrm{5}}\left(\mathrm{1}+\sqrt{\mathrm{2}}\:\right)\right) \\ $$$$\:\left({A}\:{suivre}\:\:\right)...... \\ $$

Commented by a.lgnaoui last updated on 09/Dec/22

Answered by mr W last updated on 10/Dec/22

the section (S) of both surfaces is   a circle with  center at ((4/3),(4/3),(4/3))  and radius ((4(√6))/3)

$${the}\:{section}\:\left({S}\right)\:{of}\:{both}\:{surfaces}\:{is}\: \\ $$$${a}\:{circle}\:{with} \\ $$$${center}\:{at}\:\left(\frac{\mathrm{4}}{\mathrm{3}},\frac{\mathrm{4}}{\mathrm{3}},\frac{\mathrm{4}}{\mathrm{3}}\right) \\ $$$${and}\:{radius}\:\frac{\mathrm{4}\sqrt{\mathrm{6}}}{\mathrm{3}} \\ $$

Commented by mr W last updated on 10/Dec/22

sorry, i checked again. the section is  indeed a circle. i have fixed my answer.

$${sorry},\:{i}\:{checked}\:{again}.\:{the}\:{section}\:{is} \\ $$$${indeed}\:{a}\:{circle}.\:{i}\:{have}\:{fixed}\:{my}\:{answer}. \\ $$

Commented by a.lgnaoui last updated on 10/Dec/22

thank you

$${thank}\:{you}\: \\ $$

Commented by mr W last updated on 10/Dec/22

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