Answer to Question #223830 in Chemical Engineering for Lokika

Question #223830

The solution of p^{2}+q^{2}=2 is


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

"p^2+q^2=2\\\\\n\\left(x\u2212a\\right)^2+\\left(y\u2212b\\right)^2=r^2\\:\\:\\mathrm{is\\:the\\:circle\\:equation\\:with\\:a\\:radius\\:r,\\:centered\\:at}\\:\\left(a,\\:b\\right)\\\\\n\\mathrm{Rewrite}\\:p^2+q^2=2\\:\\mathrm{in\\:the\\:form\\:of\\:the\\:standard\\:circle\\:equation}\\\\\n\\left(p-0\\right)^2+\\left(q-0\\right)^2=\\left(\\sqrt{2}\\right)^2\\\\\nTherefore\\:the\\:circle\\:properties\\:are:\\\\\n\\left(a,\\:b\\right)=\\left(0,\\:0\\right),\\:r=\\sqrt{2}"


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