{"id":5513,"date":"2010-11-15T01:46:33","date_gmt":"2010-11-15T01:46:33","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=5513"},"modified":"2022-01-12T11:14:40","modified_gmt":"2022-01-12T11:14:40","slug":"circunferencia-circunscrita-no-triangulo-abc","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=5513","title":{"rendered":"Circunfer\u00eancia circunscrita no tri\u00e2ngulo [ABC]"},"content":{"rendered":"<p><ul id='GTTabs_ul_5513' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_5513' class='GTTabs_curr'><a  id=\"5513_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_5513' ><a  id=\"5513_1\" onMouseOver=\"GTTabsShowLinks('Resolu\u00e7\u00e3o'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Resolu\u00e7\u00e3o<\/a><\/li>\n<\/ul>\n\n<div class='GTTabs_divs GTTabs_curr_div' id='GTTabs_0_5513'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p>Considere o tri\u00e2ngulo [ABC], sendo $A(-5,1)$, $B(1,3)$ e $C(3,1)$.<\/p>\n<ol>\n<li>Escreva uma equa\u00e7\u00e3o cartesiana da mediatriz do lado [AB].<\/li>\n<li>Escreva uma equa\u00e7\u00e3o cartesiana da mediatriz do lado [BC].<\/li>\n<li>Determine as coordenadas do ponto de intersec\u00e7\u00e3o das medianas determinadas (circuncentro ou centro da circunfer\u00eancia circunscrita no tri\u00e2ngulo).<\/li>\n<li>Escreva uma equa\u00e7\u00e3o da circunfer\u00eancia circunscrita ao tri\u00e2ngulo.<\/li>\n<\/ol>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_5513' onClick='GTTabs_show(1,5513)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_5513'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<ol>\n<li>As coordenadas do ponto m\u00e9dio de [AB] s\u00e3o ${{M}_{[AB]}}(\\frac{-5+1}{2},\\frac{1+3}{2})=(-2,2)$.<br \/>\nSendo ${{M}_{1}}(x,y)$ um ponto gen\u00e9rico da mediatriz de [AB], esta reta pode ser definida por: \\[\\begin{array}{*{35}{l}}<br \/>\n\\overrightarrow{{{M}_{[AB]}}{{M}_{1}}}.\\overrightarrow{AB}=0 &amp; \\Leftrightarrow\u00a0 &amp; (x+2,y-2).(6,2)=0\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; 6x+12+2y-4=0\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; y=-3x-4\u00a0 \\\\<br \/>\n\\end{array}\\]<br \/>\n\u00ad<\/li>\n<li>As coordenadas do ponto m\u00e9dio de [BC] s\u00e3o ${{M}_{[BC]}}(\\frac{1+3}{2},\\frac{3+1}{2})=(2,2)$.<br \/>\nSendo ${{M}_{2}}(x,y)$ um ponto gen\u00e9rico da mediatriz de [BC], esta reta pode ser definida por: \\[\\begin{array}{*{35}{l}}<br \/>\n\\overrightarrow{{{M}_{[BC]}}{{M}_{2}}}.\\overrightarrow{BC}=0 &amp; \\Leftrightarrow\u00a0 &amp; (x-2,y-2).(2,-2)=0\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; 2x-4-2y+4=0\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; y=x\u00a0 \\\\<br \/>\n\\end{array}\\]<br \/>\n\u00ad<\/li>\n<li>Ora,\n<p>$\\begin{matrix}<br \/>\n\\left\\{ \\begin{array}{*{35}{l}}<br \/>\ny=-3x-4\u00a0 \\\\<br \/>\ny=x\u00a0 \\\\<br \/>\n\\end{array} \\right. &amp; \\Leftrightarrow\u00a0 &amp; \\left\\{ \\begin{array}{*{35}{l}}<br \/>\nx=-3x-4\u00a0 \\\\<br \/>\ny=x\u00a0 \\\\<br \/>\n\\end{array} \\right. &amp; \\Leftrightarrow\u00a0 &amp; \\left\\{ \\begin{array}{*{35}{l}}<br \/>\nx=-1\u00a0 \\\\<br \/>\ny=-1\u00a0 \\\\<br \/>\n\\end{array} \\right.\u00a0 \\\\<br \/>\n\\end{matrix}$<\/p>\n<p>Logo, o ponto de intersec\u00e7\u00e3o das duas mediatrizes \u00e9 $E(-1,-1)$.<br \/>\n\u00ad<\/p>\n<\/li>\n<li>\n<p>O centro da circunfer\u00eancia \u00e9 $E(-1,-1)$ e o raio \u00e9 $r=\\overline{EA}=\\sqrt{{{(-5+1)}^{2}}+{{(1+1)}^{2}}}=2\\sqrt{5}$.<\/p>\n<p>Logo, essa circunfer\u00eancia pode ser definida por: ${{(x+1)}^{2}}+{{(y+1)}^{2}}=20$.<br \/>\n\u00ad<\/p>\n<\/li>\n<\/ol>\n<p><strong>Verifique a resolu\u00e7\u00e3o, manipulando a anima\u00e7\u00e3o seguinte<\/strong>:<\/p>\n<p style=\"text-align: center;\"><script src=\"https:\/\/cdn.geogebra.org\/apps\/deployggb.js\"><\/script>\r\n<div id=\"ggbApplet\" style=\"margin: 0 auto;\"><\/div>\r\n<script>\r\nvar parameters = {\r\n\"id\": \"ggbApplet\",\r\n\"width\":699,\r\n\"height\":464,\r\n\"showMenuBar\":false,\r\n\"showAlgebraInput\":false,\r\n\"showToolBar\":false,\r\n\"customToolBar\":\"0 39 | 1 501 67 , 5 19 , 72 | 2 15 45 , 18 65 , 7 37 | 4 3 8 9 , 13 44 , 58 , 47 | 16 51 64 , 70 | 10 34 53 11 , 24  20 22 , 21 23 | 55 56 57 , 12 | 36 46 , 38 49  50 , 71 | 30 29 54 32 31 33 | 17 26 62 73 , 14 68 | 25 52 60 61 | 40 41 42 , 27 28 35 , 6\",\r\n\"showToolBarHelp\":false,\r\n\"showResetIcon\":true,\r\n\"enableLabelDrags\":false,\r\n\"enableShiftDragZoom\":false,\r\n\"enableRightClick\":false,\r\n\"errorDialogsActive\":false,\r\n\"useBrowserForJS\":false,\r\n\"preventFocus\":false,\r\n\"showFullscreenButton\":true,\r\n\"language\":\"pt\",\r\n\/\/ use this instead of ggbBase64 to load a material from GeoGebraTube\r\n\/\/ 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Escreva uma equa\u00e7\u00e3o cartesiana da mediatriz do lado [AB]. Escreva uma equa\u00e7\u00e3o cartesiana da mediatriz do lado [BC]. Determine as coordenadas do&#46;&#46;&#46;<\/p>\n","protected":false},"author":1,"featured_media":19486,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[98,97,110],"tags":[422,67,111],"series":[],"class_list":["post-5513","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-11--ano","category-aplicando","category-geometria-analitica","tag-11-o-ano","tag-geometria","tag-vectores"],"views":8962,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/11\/Circunferencia_circunscrita_no_triangulo_ABC_2.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/5513","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=5513"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/5513\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/19486"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=5513"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=5513"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=5513"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=5513"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}