The solution of p^{2}+q^{2}=2 is
p2+q2=2(x−a)2+(y−b)2=r2 is the circle equation with a radius r, centered at (a, b)Rewrite p2+q2=2 in the form of the standard circle equation(p−0)2+(q−0)2=(2)2Therefore the circle properties are:(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}p2+q2=2(x−a)2+(y−b)2=r2isthecircleequationwitharadiusr,centeredat(a,b)Rewritep2+q2=2intheformofthestandardcircleequation(p−0)2+(q−0)2=(2)2Thereforethecirclepropertiesare:(a,b)=(0,0),r=2
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