{"id":4516,"date":"2013-06-29T06:35:12","date_gmt":"2013-06-29T06:35:12","guid":{"rendered":"http:\/\/www.mometrix.com\/academy\/?page_id=4516"},"modified":"2026-03-25T11:46:17","modified_gmt":"2026-03-25T16:46:17","slug":"points-of-a-circle","status":"publish","type":"page","link":"https:\/\/www.mometrix.com\/academy\/points-of-a-circle\/","title":{"rendered":"Points on a Circle"},"content":{"rendered":"\n\t\t\t<div id=\"mmDeferVideoEncompass_IYNo7DvIlCc\" style=\"position: relative;\">\n\t\t\t<picture>\n\t\t\t\t<source srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.webp\" type=\"image\/webp\">\n\t\t\t\t<source srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" type=\"image\/jpeg\"> \n\t\t\t\t<img fetchpriority=\"high\" decoding=\"async\" loading=\"eager\" id=\"videoThumbnailImage_IYNo7DvIlCc\" data-source-videoID=\"IYNo7DvIlCc\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" alt=\"Points on a Circle Video\" height=\"464\" width=\"825\" class=\"size-full\" data-matomo-title = \"Points on a Circle\">\n\t\t\t<\/picture>\n\t\t\t<\/div>\n\t\t\t<style>img#videoThumbnailImage_IYNo7DvIlCc:hover {cursor:pointer;} img#videoThumbnailImage_IYNo7DvIlCc {background-size:contain;background-image:url(\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/02\/1873-thumb-final-2.webp\");}<\/style>\n\t\t\t<script defer>\n\t\t\t  jQuery(\"img#videoThumbnailImage_IYNo7DvIlCc\").click(function() {\n\t\t\t\tlet videoId = jQuery(this).attr(\"data-source-videoID\");\n\t\t\t\tlet helpTag = '<div id=\"mmDeferVideoYTMessage_IYNo7DvIlCc\" style=\"display: none;position: absolute;top: -24px;width: 100%;text-align: center;\"><span style=\"font-style: italic;font-size: small;border-top: 1px solid #fc0;\">Having trouble? <a href=\"https:\/\/www.youtube.com\/watch?v='+videoId+'\" target=\"_blank\">Click here to watch on YouTube.<\/a><\/span><\/div>';\n\t\t\t\tlet tag = document.createElement(\"iframe\");\n\t\t\t\ttag.id = \"yt\" + videoId;\n\t\t\t\ttag.src = \"https:\/\/www.youtube-nocookie.com\/embed\/\" + videoId + \"?autoplay=1&controls=1&wmode=opaque&rel=0&egm=0&iv_load_policy=3&hd=0&enablejsapi=1\";\n\t\t\t\ttag.frameborder = 0;\n\t\t\t\ttag.allow = \"autoplay; fullscreen\";\n\t\t\t\ttag.width = this.width;\n\t\t\t\ttag.height = this.height;\n\t\t\t\ttag.setAttribute(\"data-matomo-title\",\"Points on a Circle\");\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_IYNo7DvIlCc\").html(tag);\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_IYNo7DvIlCc\").prepend(helpTag);\n\t\t\t\tsetTimeout(function(){jQuery(\"div#mmDeferVideoYTMessage_IYNo7DvIlCc\").css(\"display\", \"block\");}, 2000);\n\t\t\t  });\n\t\t\t  \n\t\t\t<\/script>\n\t\t\n<p><script>\nfunction BYH_Function() {\n  var x = document.getElementById(\"BYH\");\n  if (x.style.display === \"none\") {\n    x.style.display = \"block\";\n  } else {\n    x.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<div class=\"moc-toc hide-on-desktop hide-on-tablet\">\n<div><button onclick=\"BYH_Function()\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2024\/12\/toc2.svg\" width=\"16\" height=\"16\" alt=\"show or hide table of contents\"><\/button><\/p>\n<p>On this page<\/p>\n<\/div>\n<nav id=\"BYH\" style=\"display:none;\">\n<ul>\n<li class=\"toc-h2\"><a href=\"#Finding_Points_on_a_Circle\" class=\"smooth-scroll\">Finding Points on a Circle<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Points_on_a_Circle_Practice_Questions\" class=\"smooth-scroll\">Points on a Circle Practice Questions<\/a><\/li>\n<\/ul>\n<\/nav>\n<\/div>\n<div class=\"accordion\"><input id=\"transcript\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"transcript\">Transcript<\/label><input id=\"PQs\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQs\">Practice<\/label>\n<div class=\"spoiler\" id=\"transcript-spoiler\">\n<p>Most of us learned how to draw a circle in math class using a compass. All we needed to know was where to put the center of the circle and the measure of the radius to set on the compass. Then we held the compass steady at the center and rotated the pencil part all the way around to draw our circle. If we drew it on graph paper and looked closely, we could probably find a few specific points on our circle.<\/p>\n<p>But what if we don\u2019t have a compass? Or graph paper? Or even the radius of the circle? What if we are only given the center of the circle and one point on the circle? Can we find other points on the circle? <\/p>\n<p>The answer, of course, is yes. And that\u2019s what we\u2019re going to do today. <\/p>\n<h2><span id=\"Finding_Points_on_a_Circle\" class=\"m-toc-anchor\"><\/span>Finding Points on a Circle<\/h2>\n<p>\nLet\u2019s start by looking at the <strong>equation of a circle<\/strong>: <\/p>\n<div class=\"examplesentence\">\\((x-h)^{2}+(y-k)^{2}=r^{2}\\)<\/div>\n<p>\n&nbsp;<br \/>\nWow that\u2019s a lot of variables! The \\(x\\) and the \\(y\\) are pretty standard, but the others might not be familiar. The \\(h\\) and the \\(k\\) together make the center of the circle at the point \\((h,k)\\). The \\(r\\) is the radius of the circle. <\/p>\n<p>So if we have a circle with a center at the point \\((2,3)\\) and a radius of 5, our equation would look like this: <\/p>\n<div class=\"examplesentence\">\\((x-2)^{2}+(y-3)^{2}=25\\)<\/div>\n<p>\n&nbsp;<br \/>\nWe simply plug in the values for \\(h\\) and \\(k\\), and then square the radius for the right side of the equation. Be careful when \\(h\\) and \\(k\\) are negative, though. For instance, if the radius is 4 and the center of the circle is at \\((-4,-5)\\) and we substitute into our equation, we have to deal with subtracting negatives: <\/p>\n<div class=\"examplesentence\">\\((x-(-4))^{2}+(y-(-5))^{2}\\)\\(=4^{2}\\)<\/div>\n<p>\n&nbsp;<br \/>\nThose two negative signs in a row cancel and we end up with:<\/p>\n<div class=\"examplesentence\">\\((x+4)^{2}+(y+5)^{2}=16\\)<\/div>\n<p>\n&nbsp;<br \/>\nWhen the center of the circle is at the point \\((0,0)\\), also known as the origin, our equation gets simpler: <\/p>\n<div class=\"examplesentence\">\\((x-0)^{2}+(y-0)^{2}=r^{2}\\)<\/div>\n<p>\n&nbsp;<\/p>\n<div class=\"examplesentence\">\\(x^{2}+y^{2}=r^{2}\\)<\/div>\n<p>\n&nbsp;<br \/>\nSimple, right? But we\u2019ve gotten a bit carried away. Let\u2019s get back to our problem. If we\u2019re given the center of a circle and 1 other point, can we find 3 other points on a circle? <\/p>\n<p>Let\u2019s do an actual problem to see how we should get this done. <\/p>\n<p>Find at least 3 other points on a circle that has a point at \\((2,6)\\) and its center at \\((-2,3)\\). <\/p>\n<p>So our point is gonna be at \\((2,6)\\). And our center is at \\((-2,3)\\).<\/p>\n<p>The first thing we need to do is find the radius of the circle. We can do this by plugging in everything we know into our equation for a circle. Since we know the center is \\((-2,3)\\), we know that \\(h=-2\\) and \\(k=3\\). Since we know a point on our circle, we can use \\(x=2\\) and \\(y=6\\). We plug it all in and it looks like this.<\/p>\n<p>So the formula for a circle is:<\/p>\n<div class=\"examplesentence\">\\((x-h)^{2}+(y-k)^{2}=r^{2}\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo our \\(x\\) and \\(y\\) comes from our point we were given, and our center is our \\(h\\) and \\(k\\).<\/p>\n<div class=\"examplesentence\">\\((2-(-2))^{2}+(6-3)^{2}=r^{2}\\)<\/div>\n<p>\n&nbsp;<br \/>\nWe deal with our subtracting a negative problem to get this. \\(2-(-2)\\) is the same as \\(2+2\\), so we just replace that minus a negative with a plus sign. And then everything else stays the same.<\/p>\n<div class=\"examplesentence\">\\((2+2)^{2}+(6-3)^{2}=r^{2}\\)<\/div>\n<p>\n&nbsp;<br \/>\nThen we evaluate what\u2019s in each set of parentheses. \\(2+2=4\\), so we have :<\/p>\n<div class=\"examplesentence\">\\((4)^{2}+(3)^{2}=r^{2}\\)<\/div>\n<p>\n&nbsp;<br \/>\nWe apply our exponents. \\(4^2=16\\) and \\(3^2=9\\).<\/p>\n<div class=\"examplesentence\">\\(16+9=r^{2}\\)<\/div>\n<p>\n&nbsp;<br \/>\nAnd then we\u2019re gonna add them and solve for \\(r\\). So, \\(16+9=25\\).<\/p>\n<div class=\"examplesentence\">\\(25=r^2\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>We\u2019ll square root both sides and get that \\(r=5\\). <\/p>\n<p>Now what? How can we use this information to find more points on our circle? We\u2019re going to use a different kind of compass to do that! Let\u2019s get out a piece of graph paper and plot what we know so far. <\/p>\n<p>Since we know our radius, we can travel north, south, east, and west from that point exactly 5 units to find more points! Moving along the compass points is easy on our graph paper, and since the radius is the distance from the center of the circle to all the points on the circle, we know we\u2019ll end up on our circle when we travel 5 units.<\/p>\n<p>Look at all those points we found! Going north five units we found \\((-2,8)\\), going west we found \\((-7,3)\\), going south we found \\((-2,-2)\\), and going east we found \\((3,3)\\)! 4 points! One more than the problem asked us to find.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/09\/Points-on-a-circle-1-e1631128126789.png\" alt=\"For a circle centered at (-2, 3) pointing out 5 points in the four compass directions lets you find the points (-2,8), (3,3), (-2,-2), and (-7,3)\" width=\"869\" height=\"845\" class=\"aligncenter size-full wp-image-90862\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/09\/Points-on-a-circle-1-e1631128126789.png 869w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/09\/Points-on-a-circle-1-e1631128126789-300x292.png 300w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/09\/Points-on-a-circle-1-e1631128126789-768x747.png 768w\" sizes=\"auto, (max-width: 869px) 100vw, 869px\" \/><\/p>\n<p>And we\u2019re not done yet! We can find even more points if we want, since the given point isn\u2019t one of our 4 \u201ccompass points.\u201d If we look to see how far over and up that point is from the center, we can find more points that are the same L-shaped distance away. Here we can see that \\((2,6)\\) is 4 units to the right and 3 units up from the center of the circle. So I can draw L-shapes from the center that move 4 units left or right and then 3 units up and down to find more points.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/09\/Points-on-a-circle-2-e1631128781317.png\" alt=\"Moving left and right 4 units and up and down 3 units from the center (-2,3) allows us to find four more points\" width=\"853\" height=\"824\" class=\"aligncenter size-full wp-image-90868\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/09\/Points-on-a-circle-2-e1631128781317.png 853w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/09\/Points-on-a-circle-2-e1631128781317-300x290.png 300w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/09\/Points-on-a-circle-2-e1631128781317-768x742.png 768w\" sizes=\"auto, (max-width: 853px) 100vw, 853px\" \/><\/p>\n<p>We found 3 more! On the left (or west) side we found \\((-6,6)\\) and \\((-6,0)\\) and below our given point we found \\((2,0)\\). Altogether we now have 8 points on our circle including the given one! And if we\u2019re willing to do some heavier math, we can find any of the infinite number of points on our circle. Since we\u2019re on such a roll, let\u2019s try that too! <\/p>\n<p>First we need to establish our domain so we know what \\(x\\)-values we can pick. The leftmost \\(x\\)-value on our circle is our \u201cwest\u201d point at -7. Our rightmost point is our \u201ceast\u201d point at +3. So our domain is \\(x\\geq -7\\) and \\(x\\leq 3\\).<\/p>\n<div class=\"examplesentence\">\\(\\text{Domain: } {{x\\mid-7\\leq x\\leq 3}}\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo we can pick any \\(x\\)-value from -7 to 3 to find its corresponding \\(y\\)-values on the circle. Yes, it&#8217;s plural because there will be 2. <\/p>\n<p>So let\u2019s pick \\(x=-4\\) from our domain. Now all we have to do is plug that into our equation for this circle and solve for \\(y\\).<\/p>\n<p>So we\u2019re using the \\(x\\)-value -4. And remember, our center is at the point \\((-2,3)\\). So now all we&#8217;re gonna do is plug that into our equation for the circle. So remember, that\u2019s:<\/p>\n<div class=\"examplesentence\">\\((x-h)^{2}+(y-k)^{2}=r^{2}\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo our \\(x\\)-value is -4, minus \\(h\\) is -2, plus \\(y\\) (we don\u2019t know what that is, that\u2019s what we\u2019re trying to solve for) minus \\(k\\), which is 3, squared, and we forgot to write this, but our \\(r\\), remember from earlier is 5, so is equal to 5 squared.<\/p>\n<div class=\"examplesentence\">\\((-4(-2))^{2}+(y-3)^{2}=r^{3}\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo \\(-4-(-2)\\), that\u2019s the same as saying \\(-4+2\\), so let\u2019s rewrite it that way.<\/p>\n<div class=\"examplesentence\">\\((-4+2)^{2}+(y-3)^{2}=(5)^{2}\\)<\/div>\n<p>\n&nbsp;<br \/>\nThen we\u2019ll simplify \\(-4+2\\) to \\(-2\\), so:<\/p>\n<div class=\"examplesentence\">\\((-2)^{2}+(y-3)^{2}=(5)^{2}\\)<\/div>\n<p>\n&nbsp;<br \/>\nAnd now we\u2019re gonna simplify our exponents. \\((-2)^2=4\\) plus (we\u2019ll leave this part alone for right now, we\u2019ll come back to that later) \\((y-3)^2\\) equals, \\(5^2=5\\).<\/p>\n<div class=\"examplesentence\">\\(4+(y-3)^{2}=25\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow we\u2019ll subtract 4 from both sides. This gives us:<\/p>\n<div class=\"examplesentence\">\\((y-3)^{2}=21\\)<\/div>\n<p>\n&nbsp;<br \/>\nAnd we\u2019ll take the square root of both sides. That leaves us with:<\/p>\n<div class=\"examplesentence\">\\(y-3=\\pm \\sqrt{21}\\)<\/div>\n<p>\n&nbsp;<br \/>\nAnd all we have to do is add 3 to both sides. So our answers are:<\/p>\n<div class=\"examplesentence\">\\(y=\\pm \\sqrt{21}+3\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo our two values are \\(\\sqrt{21}+3\\) and \\(-\\sqrt{21}+3\\). If we\u2019re plotting those points on a graph, we can estimate their value by using a calculator to find that our \\(y\\)-values are approximately 7.58 and -1.58. So we can graph those too!<\/p>\n<p>Now we\u2019re up to 10 points on our circle! Can you find any more? Pause this video now and try. The answer for all the integer values of \\(x\\) will appear shortly after unpausing. <\/p>\n<p>Here are the other points on the circle with integer \\(x\\)-values: <\/p>\n<div class=\"examplesentence\">\\((-5,-1)(-5,7)\\)<br \/>\n\\((-3,2\\sqrt{6}+3)(-3,-2\\sqrt{6}+3)\\)<br \/>\n\\((-1,2\\sqrt{6}+3)(-1,-2\\sqrt{6}+3)\\)<br \/>\n\\((0,\\sqrt{21}+3)(0,-\\sqrt{21}+3)\\)<br \/>\n\\((1,-1)(1,7)\\)<\/div>\n<p>\n&nbsp;<br \/>\nI hope this video over finding points on a circle was helpful. Thanks for watching, and happy studying! <\/p>\n<div style=\"text-align: center;\"><a class=\"ylist\" href=\"https:\/\/www.mometrix.com\/academy\/the-diameter-radius-and-circumference-of-circles\/\">Circumference of a Circle<\/a> | <a class=\"ylist\" href=\"https:\/\/www.mometrix.com\/academy\/circle-equations\/\">Circle Equations<\/a><\/div>\n<p>&nbsp;<\/p>\n<ul class=\"citelist\">\n<li><a href=\"https:\/\/www.mathopenref.com\/coordgeneralcircle.html\"target=\"_blank\">\u201cGeneral Form of Equation of a Circle &#8211; Math Open Reference.\u201d n.d.<\/a><\/li>\n<\/ul>\n<\/div>\n<div class=\"spoiler\" id=\"PQs-spoiler\">\n<h2 style=\"text-align:center\"><span id=\"Points_on_a_Circle_Practice_Questions\" class=\"m-toc-anchor\"><\/span>Points on a Circle Practice Questions<\/h2>\n\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #1:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nWhich of the following sets of points are on a circle that has its center at \\((3,6)\\) and a radius of 13 units?<\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-1-1\">\\((-8,-18)\\) and \\((2,6)\\)<\/div><div class=\"PQ\"  id=\"PQ-1-2\">\\((-3,5)\\) and \\((4,8)\\)<\/div><div class=\"PQ correct_answer\"  id=\"PQ-1-3\">\\((-2,-6)\\) and \\((8,18)\\)<\/div><div class=\"PQ\"  id=\"PQ-1-4\">\\((-4,6)\\) and \\((16,6)\\)<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-1\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-1\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-1-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>The equation of a circle in standard form is \\((x-h)^2+(y-k)^2=r^2\\), where \\((h,k)\\) is the center, and \\(r\\) is the radius of the circle. The ordered pair \\((x,y)\\) represents any point on the circle. <\/p>\n<p>Substituting our center and radius into the equation of a circle, we have:<\/p>\n<p style=\"text-align: center; line-height: 45px\">\n\\((x-3)^2+(y-6)^2=13^2\\)<br \/>\n\\((x-3)^2+(y-6)^2=169\\)<\/p>\n<p>An ordered pair \\((x,y)\\) is a point on the circle if it satisfies the equation of the circle. Substituting \\((-2,-6)\\) into the equation of the circle, we get:<\/p>\n<p style=\"text-align: center; line-height: 45px\">\n\\((-2-3)^2+(-6-6)^2=169\\)<br \/>\n\\((-5)^2+(-12)^2=169\\)<br \/>\n\\(25+144=169\\)<br \/>\n\\(169=169\\)<\/p>\n<p>Since the ordered pair produces a true statement when substituting it into the equation of the circle, it satisfies the equation, so it is a point on the circle.<\/p>\n<p>Substituting \\((8,18)\\) into our equation, we get:<\/p>\n<p style=\"text-align: center; line-height: 45px\">\n\\((8-3)^2+(18-6)^2=169\\)<br \/>\n\\((5)^2+(12)^2=169\\)<br \/>\n\\(25+144=169\\)<br \/>\n\\(169=169\\)<\/p>\n<p>Since the ordered pair \\((8,18)\\) produces a true statement when substituting it into the equation of the circle, it satisfies the equation, so it is also a point on the circle.<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-1-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-1-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #2:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nWhich of the following set of points are on a circle that has its center at \\((-2,4)\\) and contains the point \\((-2,0)\\)?<\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ correct_answer\"  id=\"PQ-2-1\">\\((0,4-2\\sqrt{3})\\) and \\((0,4+2\\sqrt{3})\\)<\/div><div class=\"PQ\"  id=\"PQ-2-2\">\\((0,4-2\\sqrt{5})\\) and \\((0,4+2\\sqrt{5})\\)<\/div><div class=\"PQ\"  id=\"PQ-2-3\">\\((2,0)\\) and \\((6,4)\\)<\/div><div class=\"PQ\"  id=\"PQ-2-4\">\\((-2,4)\\) and \\((6,0)\\)<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-2\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-2\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-2-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>Since we are given the center and a point on the circle, we can find its radius.<\/p>\n<p>The equation of a circle in standard form is \\((x-h)^2+(y-k)^2=r^2\\), where \\((h,k)\\) is the center, and \\(r\\) is the radius of the circle. The ordered pair \\((x,y)\\) represents any point on the circle. <\/p>\n<p>Substituting our center \\((-2,4)\\) into the equation of the circle, we have:<\/p>\n<p style=\"text-align: center; line-height: 45px\">\n\\((x-(-2))^2+(y-4)^2=r^2\\)<br \/>\n\\((x+2)^2+(y-4)^2=r^2\\)<\/p>\n<p>Now entering the given point \\((-2,0)\\) into the equation for the circle, we have:<\/p>\n<p style=\"text-align: center; line-height: 45px\">\n\\((-2+2)^2+(0-4)^2=r^2\\)<br \/>\n\\((-2+2)^2+(-4)^2=r^2\\)<br \/>\n\\((0)^2+(-4)^2=r^2\\)<br \/>\n\\(16=r^2\\)<br \/>\n\\(\\sqrt{16}=\\sqrt{r^2}\\)<br \/>\n\\(r=4\\)<\/p>\n<p>While \\(r=\\pm4\\), we only use 4 since the radius has a non-negative length. <\/p>\n<p>Substituting our center and radius into the equation of a circle, we have:<\/p>\n<p style=\"text-align: center; line-height: 45px\">\n\\((x+2)^2+(y-4)^2=4^2\\)<br \/>\n\\((x+2)^2+(y-4)^2=16\\)<\/p>\n<p>In this case, since the given point is directly below the center of the circle, we could also find the radius by counting 4 units from the center to the given point (or 4 units from the given point to the center) as shown in the figure below.<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2026\/02\/Points-of-a-Circle-Example-1.svg\" alt=\"Red circle with center at (-2, 4) and radius 4 on a coordinate plane; radius is shown as a dashed line from the center to the edge at (-2, 0).\" width=\"392.15\" height=\"368.9\" class=\"aligncenter size-full wp-image-287017\"  role=\"img\" \/><\/p>\n<p>An ordered pair \\((x,y)\\) is a point on the circle if it satisfies the equation of the circle. Substituting \\((0,4-2\\sqrt{3})\\) into the equation of the circle, we get:<\/p>\n<p style=\"text-align: center; line-height: 45px\">\n\\((0+2)^2+(4-2\\sqrt{3}-4)^2=16\\)<br \/>\n\\((2)^2+(-2\\sqrt{3})^2=16\\)<br \/>\n\\(4+4\\cdot3=16\\)<br \/>\n\\(4+12=16\\)<br \/>\n\\(16=16\\)<\/p>\n<p>Since the ordered pair produces a true statement when substituting it into the equation of the circle, it satisfies the equation, so it is a point on the circle.<\/p>\n<p>Substituting \\((0,4+2\\sqrt{3})\\) into the equation of the circle, we get:<\/p>\n<p style=\"text-align: center; line-height: 45px\">\n\\((0+2)^2+(4+2\\sqrt{3}-4)^2=16\\)<br \/>\n\\((2)^2+(2\\sqrt{3})^2=16\\)<br \/>\n\\(4+4\\cdot3=16\\)<br \/>\n\\(4+12=16\\)<br \/>\n\\(16=16\\)<\/p>\n<p>Since the ordered pair \\((0,4+2\\sqrt{3})\\) produces a true statement when substituting it into the equation of the circle, it satisfies the equation, so it is also a point on the circle.<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-2-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-2-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #3:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nWhich are the leftmost and rightmost points that are on a circle that has a center at \\((4,-1)\\) and contains the point \\((8,2)\\)?<\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-3-1\">\\((4,-6)\\) and \\((4,4)\\)<\/div><div class=\"PQ correct_answer\"  id=\"PQ-3-2\">\\((-1,-1)\\) and \\((9,-1)\\)<\/div><div class=\"PQ\"  id=\"PQ-3-3\">\\((-2,-1)\\) and \\((10,-1)\\)<\/div><div class=\"PQ\"  id=\"PQ-3-4\">\\((0,-2)\\) and \\((8,-2)\\)<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-3\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-3\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-3-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>Since we are given the center and a point on the circle, we can find its radius. Then we can use the radius to find the circle\u2019s leftmost and rightmost points.<\/p>\n<p>The equation of a circle in standard form is \\((x-h)^2+(y-k)^2=r^2\\), where \\((h,k)\\) is the center, and \\(r\\) is the radius of the circle. The ordered pair \\((x,y)\\) represents any point on the circle.<br \/>\nSubstituting our center \\((4,-1)\\) into the equation of the circle, we have:<\/p>\n<p style=\"text-align: center; line-height: 45px\">\n\\((x-4)^2+(y-(-1))^2=r^2\\)<br \/>\n\\((x-4)^2+(y+1)^2=r^2\\)>\/p><\/p>\n<p>Now entering the given point \\((8,2)\\) into the equation for the circle, we have:<\/p>\n<p style=\"text-align: center; line-height: 45px\">\n\\((8-4)^2+(2+1)^2=r^2\\)<br \/>\n\\((4)^2+(3)^2=r^2\\)<br \/>\n\\(16+9=r^2\\)<br \/>\n\\(25=r^2\\)<br \/>\n\\(\\sqrt{25}=\\sqrt{r^2}\\)<br \/>\n\\(r=5\\)<\/p>\n<p>While \\(r=\\pm5\\), we only use 5 since the radius has a non-negative length.<\/p>\n<p>On the coordinate plane, we can use the value of \\(r\\) to count a radius length of 5 units horizontally to the left and right (or west and east) from the center of the circle to find the furthest most left and right points, respectively. <\/p>\n<p><img decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2026\/02\/Points-of-a-Circle-Example-2.svg\" alt=\"A red circle is centered at (4, -1) with points labeled (-1, -1), (8, 2), and (9, -1) on or near its edge, on a grid with marked axes.\" width=\"474.3\" height=\"395.25\" class=\"aligncenter size-full wp-image-287020\"  role=\"img\" \/><\/p>\n<p>Counting a radius length of 5 units to the west from the center produces the leftmost point of \\((-1,-1)\\) on the circle. Counting a radius length of 5 units to the east from the center produces rightmost point of \\((9,-1)\\).<\/p>\n<p>Counting a radius length of 5 units from the center to the west and east always produces the leftmost and rightmost points on the circle. Below are 2 additional points that are graphed on the circle. While both points are also 5 units from the center, notice that neither are the leftmost or rightmost points. <\/p>\n<p><img decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2026\/02\/Points-of-a-Circle-Example-3.svg\" alt=\"Red circle centered at (4, -1) with labeled points at (0,2), (8,2), (9,1), (8.6,-3), and (-1,-1); radius is 5; dashed arrowed diameter shown.\" width=\"474.3\" height=\"395.25\" class=\"aligncenter size-full wp-image-287014\"  role=\"img\" \/><\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-3-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-3-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #4:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nYou tie your pet with a rope to a stake that is mounted to the ground. When the rope is fully extended, your pet can walk a circular path around the stake. If the rope is 10 feet long and the stake is mounted at the origin of the coordinate plane, which of the following points are on the path that your pet takes?<\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-4-1\">\\((0,20)\\) and \\((10,10)\\)<\/div><div class=\"PQ\"  id=\"PQ-4-2\">\\((0,-5)\\) and \\((2,5)\\)<\/div><div class=\"PQ\"  id=\"PQ-4-3\">\\((-4,-6)\\) and \\((4,6)\\)<\/div><div class=\"PQ correct_answer\"  id=\"PQ-4-4\">\\((-8,6)\\) and \\((6,8)\\)<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-4\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-4\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-4-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>The equation of a circle in standard form is \\((x-h)^2+(y-k)^2=r^2\\), where \\((h,k)\\) is the center of the circle, and \\(r\\) is the radius of the circle. The ordered pair \\((x,y)\\) represents any point on the circle. Since the stake is at the origin of the coordinate plane, it is a point that has coordinates \\((0,0)\\). When the rope is fully extended, it forms the radius for the circular path your pet walks, so \\(r=10\\).<\/p>\n<p>Substituting our center and radius into the equation of a circle, we have:<\/p>\n<p style=\"text-align: center; line-height: 45px\">\n\\((x-0)^2+(y-0)^2=10^2\\)<br \/>\n\\(x^2+y^2=100\\)<\/p>\n<p>An ordered pair \\((x,y)\\) is a point on the circle if it satisfies the equation of the circle. Substituting \\((-8,6)\\) into the equation of the circle, we get:<\/p>\n<p style=\"text-align: center; line-height: 45px\">\n\\((-8)^2+(6)^2=100\\)<br \/>\n\\(64+36=100\\)<br \/>\n\\(100=100\\)<\/p>\n<p>Since the ordered pair produces a true statement when substituting it into the equation of the circle, it satisfies the equation, so it is a point on the circle.<\/p>\n<p>Substituting \\((6,8)\\) into our equation, we get:<\/p>\n<p style=\"text-align: center; line-height: 45px\">\n\\((6)^2+(8)^2=100\\)<br \/>\n\\(36+64=100\\)<br \/>\n\\(100=100\\)<\/p>\n<p>Since the ordered pair \\((6,8)\\) produces a true statement when substituting it into the equation of the circle, it satisfies the equation, so it is also a point on the circle.<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-4-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-4-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #5:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nThe ends of the spokes for a bicycle tire are attached to the center of the tire\u2019s hub and its rim. The tire contains 20 spokes. If the hub is centered at the origin of the coordinate plane and the end of one of its spokes is at the point \\((6,-2)\\) in the coordinate plane, on which set of points could the ends of two of the other 19 spokes lie?<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2026\/02\/Points-of-a-Circle-Example-4.svg\" alt=\"A bicycle wheel with labeled coordinates at the center (0,0) and a point on the rim (6,-2).\" width=\"421.6\" height=\"375.1\" class=\"aligncenter size-full wp-image-287023\"  role=\"img\" \/><\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-5-1\">\\((-4,10)\\) and \\((2,20)\\)<\/div><div class=\"PQ\"  id=\"PQ-5-2\">\\((0,40)\\) and \\((40,0)\\)<\/div><div class=\"PQ correct_answer\"  id=\"PQ-5-3\">\\((-2,6)\\) and \\((5,\\sqrt{15})\\)<\/div><div class=\"PQ\"  id=\"PQ-5-4\">\\((2,10)\\) and \\((4,-2\\sqrt{10})\\)<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-5\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-5\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-5-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>Since the center of the hub of the bicycle is at the origin of the coordinate plane, it is a point that has coordinates \\((0,0)\\). We can use the origin and the given point on the circle to find its radius. <\/p>\n<p>The equation of a circle in standard form is \\((x-h)^2+(y-k)^2=r^2\\), where \\((h,k)\\) is the center of the circle, and \\(r\\) is the radius of the circle. The ordered pair \\((x,y)\\) represents any point on the circle.<\/p>\n<p>Substituting our center \\((0,0)\\) into the equation of the circle, we have:<\/p>\n<p style=\"text-align: center; line-height: 45px\">\n\\((x-0)^2+(y-0)^2=r^2\\)<br \/>\n\\(x^2+y^2=r^2\\)<\/p>\n<p>Now entering the given point \\((6,-2)\\) into the equation for the circle, we have:<\/p>\n<p style=\"text-align: center; line-height: 45px\">\n\\((6)^2+(-2)^2=r^2\\)<br \/>\n\\(36+4=r^2\\)<br \/>\n\\(\\sqrt{40}=\\sqrt{r^2}\\)<br \/>\n\\(r=\\sqrt{40}\\) or \\(2\\sqrt{10}\\)<\/p>\n<p>While \\(r=\\pm2\\sqrt{10}\\), we only use \\(2\\sqrt{10}\\) since the radius has a non-negative length.<\/p>\n<p>Substituting our center and radius into the equation, we have:<\/p>\n<p style=\"text-align: center; line-height: 45px\">\n\\(x^2+y^2=(2\\sqrt{10})^2\\)<br \/>\n\\(x^2+y^2=4\\cdot10\\)<br \/>\n\\(x^2+y^2=40\\)<\/p>\n<p>An ordered pair \\((x,y)\\) is a point on the circle if it satisfies the equation of the circle. Substituting \\((-2,6)\\) into the equation of the circle, we get:<\/p>\n<p style=\"text-align: center; line-height: 45px\">\n\\(x^2+y^2=40\\)<br \/>\n\\((-2)^2+( 6)^2=40\\)<br \/>\n\\(4+36=40\\)<br \/>\n\\(40=40\\)<\/p>\n<p>Since the ordered pair produces a true statement when substituting it into the equation of the circle, it satisfies the equation, so it is a point on the circle.<\/p>\n<p>Substituting \\((5,\\sqrt{15})\\) into the equation of the circle, we get:<\/p>\n<p style=\"text-align: center; line-height: 45px\">\n\\((5)^2+(\\sqrt{15})^2=40\\)<br \/>\n\\(25+15=40\\)<br \/>\n\\(40=40\\)<\/p>\n<p>Since the ordered pair \\((5,\\sqrt{15})\\) produces a true statement when substituting it into the equation of the circle, it satisfies the equation, so it is also a point on the circle.<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-5-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-5-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div><\/div>\n<\/div>\n\n<div class=\"home-buttons\">\n<p><a href=\"https:\/\/www.mometrix.com\/academy\/geometry\/\">Return to Geometry Videos<\/a><\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Return to Geometry Videos<\/p>\n","protected":false},"author":1,"featured_media":100345,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"open","template":"","meta":{"footnotes":""},"class_list":{"0":"post-4516","1":"page","2":"type-page","3":"status-publish","4":"has-post-thumbnail","6":"page_category-circle-video","7":"page_category-math-advertising-group","8":"page_type-video","9":"content_type-practice-questions","10":"subject_matter-math"},"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/4516","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/comments?post=4516"}],"version-history":[{"count":6,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/4516\/revisions"}],"predecessor-version":[{"id":238852,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/4516\/revisions\/238852"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media\/100345"}],"wp:attachment":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media?parent=4516"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}