O centro da circunfer\u00eancia \u00e9 o ponto m\u00e9dio do segmento [AB]: $M(\\frac{2-2}{2},\\frac{1+3}{2})=(0,2)$.\nO raio da circunfer\u00eancia \u00e9 $r=\\frac{\\overline{AB}}{2}=\\frac{1}{2}\\sqrt{{{(-2-2)}^{2}}+{{(3-1)}^{2}}}=\\sqrt{5}$.<\/p>\n
Logo, uma equa\u00e7\u00e3o dessa circunfer\u00eancia \u00e9 ${{x}^{2}}+{{(y-2)}^{2}}=5$.<\/p>\n
ALTERNATIVA<\/strong>:
\nDesignando por $P(x,y)$ um ponto gen\u00e9rico dessa circunfer\u00eancia, tem-se $\\overrightarrow{AP}.\\overrightarrow{BP}=0$,\u00a0 pois os vetores s\u00e3o perpendiculares ou um deles \u00e9 nulo (ver anima\u00e7\u00e3o).<\/p>\nLogo, a circunfer\u00eancia pedida pode ser definida por: \\[\\begin{array}{*{35}{l}}
\n\\overrightarrow{AP}.\\overrightarrow{BP}=0 & \\Leftrightarrow\u00a0 & (x-2,y-1)(x+2,y-3)=0\u00a0 \\\\
\n{} & \\Leftrightarrow\u00a0 & ({{x}^{2}}-4)+({{y}^{2}}-4y+3)=0\u00a0 \\\\
\n{} & \\Leftrightarrow\u00a0 & {{x}^{2}}-4+{{(y-2)}^{2}}-4+3=0\u00a0 \\\\
\n{} & \\Leftrightarrow\u00a0 & {{x}^{2}}+{{(y-2)}^{2}}=5\u00a0 \\\\
\n\\end{array}\\]<\/p>\n<\/li>\n<\/ol>\n
\u00ad
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