Question #223830

The solution of p^{2}+q^{2}=2 is


1
Expert's answer
2021-08-16T02:20:00-0400

p2+q2=2(xa)2+(yb)2=r2isthecircleequationwitharadiusr,centeredat(a,b)Rewritep2+q2=2intheformofthestandardcircleequation(p0)2+(q0)2=(2)2Thereforethecirclepropertiesare:(a,b)=(0,0),r=2p^2+q^2=2\\ \left(x−a\right)^2+\left(y−b\right)^2=r^2\:\:\mathrm{is\:the\:circle\:equation\:with\:a\:radius\:r,\:centered\:at}\:\left(a,\:b\right)\\ \mathrm{Rewrite}\:p^2+q^2=2\:\mathrm{in\:the\:form\:of\:the\:standard\:circle\:equation}\\ \left(p-0\right)^2+\left(q-0\right)^2=\left(\sqrt{2}\right)^2\\ Therefore\:the\:circle\:properties\:are:\\ \left(a,\:b\right)=\left(0,\:0\right),\:r=\sqrt{2}


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