Question Number 149428 by mr W last updated on 06/Aug/21
Commented by mr W last updated on 06/Aug/21
$${three}\:{non}−{collinear}\:{points}\:{in}\:{space}\: \\ $$$${are}\:{given}\:{with}\:{coordinates} \\ $$$${A}\left({x}_{\mathrm{1}} ,{y}_{\mathrm{1}} ,{z}_{\mathrm{1}} \right) \\ $$$${B}\left({x}_{\mathrm{2}} ,{y}_{\mathrm{2}} ,{z}_{\mathrm{2}} \right) \\ $$$${C}\left({x}_{\mathrm{3}} ,{y}_{\mathrm{3}} ,{z}_{\mathrm{3}} \right) \\ $$$${find}\:{the}\:{coordinates}\:{of}\:{the}\:{center}\:{S} \\ $$$${and}\:{the}\:{radius}\:{R}\:{of}\:{the}\:{circumcircle}. \\ $$
Commented by ajfour last updated on 06/Aug/21
$${sir},\:\left({apart}\:{from}\:{this}\right);\:{sir}\:{try}\:{plotting} \\ $$$${z}\left({x},{y}\right)=\left\{\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{3}} +\left({y}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{3}} \right\}^{\mathrm{2}} \\ $$$${see}\:{all}\:{views},\:{change}\:{scale}, \\ $$$${wonderful}\:\left({i}\:{find}\right)… \\ $$
Commented by ajfour last updated on 06/Aug/21
Answered by ajfour last updated on 06/Aug/21