Question

The solution of p^{2}+q^{2}=2 is

Answer 1

$p^2+q^2=2\\ \left(x−a\right)^2+\left(y−b\right)^2=r^2\:\:\mathrm{is\:the\:circle\:equation\:with\:a\:radius\:r,\:centered\:at}\:\left(a,\:b\right)\\ \mathrm{Rewrite}\:p^2+q^2=2\:\mathrm{in\:the\:form\:of\:the\:standard\:circle\:equation}\\ \left(p-0\right)^2+\left(q-0\right)^2=\left(\sqrt{2}\right)^2\\ Therefore\:the\:circle\:properties\:are:\\ \left(a,\:b\right)=\left(0,\:0\right),\:r=\sqrt{2}$