Question

solve p
$p^{2} + q^{2} = 1$
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Answer 1

Answer:

$p^2+q^2=1\\= \left(x -a\right)^2+\left(y \\-b\right)^2=r^2\:\:\mathrm{is\:the\:circle\:equation\:with\:a\:radius\:r,\:centered\:at}\:\left(a,\:b\right)\\= \left(p-0\right)^2+\left(q-0\right)^2=1^2\\= \left(a,\:b\right)=\left(0,\:0\right),\:r=1\\So, \\p = 1 and \\q = 1$

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