{"id":37914,"date":"2026-02-06T17:01:58","date_gmt":"2026-02-06T17:01:58","guid":{"rendered":"https:\/\/mp.moonpreneur.com\/math-corner\/?p=37914"},"modified":"2026-03-06T17:38:21","modified_gmt":"2026-03-06T17:38:21","slug":"problems-on-circle-and-tangent","status":"publish","type":"post","link":"https:\/\/mp.moonpreneur.com\/math-corner\/problems-on-circle-and-tangent\/","title":{"rendered":"Geometry Practice: Essential Circle and Tangent Problems for Exams"},"content":{"rendered":"\t\t<div data-elementor-type=\"wp-post\" data-elementor-id=\"37914\" class=\"elementor elementor-37914\" data-elementor-post-type=\"post\">\n\t\t\t\t\t\t<div class=\"elementor-inner\">\n\t\t\t\t<div class=\"elementor-section-wrap\">\n\t\t\t\t\t\t\t\t\t<section class=\"has_eae_slider elementor-section elementor-top-section elementor-element elementor-element-fcfec80 elementor-section-boxed elementor-section-height-default elementor-section-height-default\" data-id=\"fcfec80\" data-element_type=\"section\">\n\t\t\t\t\t\t<div class=\"elementor-container elementor-column-gap-default\">\n\t\t\t\t\t\t\t<div class=\"elementor-row\">\n\t\t\t\t\t<div class=\"has_eae_slider elementor-column elementor-col-100 elementor-top-column elementor-element elementor-element-92390b3\" data-id=\"92390b3\" data-element_type=\"column\">\n\t\t\t<div class=\"elementor-column-wrap elementor-element-populated\">\n\t\t\t\t\t\t\t<div class=\"elementor-widget-wrap\">\n\t\t\t\t\t\t<div class=\"elementor-element elementor-element-0fa4990 elementor-widget elementor-widget-text-editor\" data-id=\"0fa4990\" data-element_type=\"widget\" data-widget_type=\"text-editor.default\">\n\t\t\t\t<div class=\"elementor-widget-container\">\n\t\t\t\t\t\t\t\t<div class=\"elementor-text-editor elementor-clearfix\">\n\t\t\t\t<p><span style=\"font-weight: 400;\">If you\u2019ve ever looked at a geometry problem involving circles and felt like you were going in&#8230; well, circles&#8230; You aren&#8217;t alone. For many SAT students, the geometry section can feel like a collection of abstract rules, but when it comes to <\/span><b>tangents<\/b><span style=\"font-weight: 400;\">, there are actually some very reliable &#8220;cheat codes&#8221; you can use to ace your exam.<\/span><\/p><p><span style=\"font-weight: 400;\">Think of a <\/span><b>tangent<\/b><span style=\"font-weight: 400;\"> as a straight line that gives a circle a &#8220;high-five&#8221;\u2014it touches the circumference at exactly one point and then keeps on going without ever crossing into the circle&#8217;s territory.<\/span><\/p><h3><span style=\"color: #ff0000;\"><b>The &#8220;Big Three&#8221; Rules You Need to Know<\/b><\/span><\/h3><p><span style=\"font-weight: 400;\">To solve most SAT circle problems, you really only need to keep three major theorems in your back pocket:<\/span><\/p><ol><li><b>The 90-Degree Rule:<\/b><span style=\"font-weight: 400;\"> This is the most common trick on the SAT. Whenever a radius meets a tangent at the point of contact, they form a perfect <\/span><b>90-degree angle<\/b><span style=\"font-weight: 400;\">. If you see a tangent, immediately look for the radius. Creating that right angle often reveals a hidden right triangle, allowing you to use the Pythagorean theorem to find missing lengths.<\/span><\/li><\/ol><ol start=\"2\"><li><b>The &#8220;Hat&#8221; Rule (Tangents from a Common Point):<\/b><span style=\"font-weight: 400;\"> If you draw two tangents from the same external point to the same circle, those two lines are <\/span><b>exactly the same length<\/b><span style=\"font-weight: 400;\">. It looks a bit like the circle is wearing a party hat. Because those two sides are equal, you often end up with an <\/span><b>isosceles triangle<\/b><span style=\"font-weight: 400;\">, which is a huge clue for solving angle problems.<\/span><\/li><\/ol><ol start=\"2\"><li><b>The Alternate Segment Theorem:<\/b><span style=\"font-weight: 400;\"> This is for the &#8220;hard&#8221; questions. It states that the angle between a tangent and a chord is equal to the angle in the alternate segment. If you can spot this relationship, you can often find a missing angle in seconds that other students might spend minutes trying to calculate.<\/span><\/li><\/ol><h4><span style=\"color: #008000;\"><b>Your SAT Game Plan<\/b><\/span><\/h4><p><span style=\"font-weight: 400;\">When you encounter a complex circle diagram, don&#8217;t panic. Follow these three steps:<\/span><\/p><ul><li><b>Locate the key parts:<\/b><span style=\"font-weight: 400;\"> Identify where the center is, where the tangents are, and where they touch (the point of tangency).<\/span><\/li><li><b>Hunt for 90-degree angles:<\/b><span style=\"font-weight: 400;\"> Draw in a radius to the point of tangency if it isn&#8217;t already there.<\/span><\/li><li><b>Check for symmetry:<\/b><span style=\"font-weight: 400;\"> If there are two tangents meeting at a point, remember they are equal in length and often form a <\/span><b>kite<\/b><span style=\"font-weight: 400;\"> shape with the radii, where the angles total 360 degrees<\/span><\/li><\/ul><h4><span style=\"color: #008000;\"><b>What is a circle?<\/b><\/span><\/h4><p><span style=\"font-weight: 400;\">In geometry, a circle is the collection of all the points in a plane that are at a fixed distance from a fixed point in the plane. The fixed point is called the centre, and the fixed distance is called the radius of the circle.<\/span><\/p><h5><span style=\"color: #000080;\"><strong>Key Circle Theorems &amp; Properties<\/strong><\/span><\/h5><ul><li><strong>Tangent-Radius Theorem:<\/strong> A tangent to a circle is perpendicular to the radius at the point of contact (90\u00b0 angle).<\/li><li><strong>Two-Tangent Theorem:<\/strong> Tangents drawn from an external point to a circle are equal in length.<\/li><li><strong> Alternate Segment Theorem:<\/strong> The angle between a tangent and a chord equals the angle in the alternate segment.<\/li><\/ul><h5><span style=\"color: #000080;\"><strong> Angle Properties:<\/strong><\/span><\/h5><ul><li>Angle at the center is twice the angle at the circumference.<\/li><li>Angles in the same segment are equal.<\/li><li>Angle in a semicircle is 90\u00b0.<\/li><\/ul><h3><span style=\"color: #008000;\"><b>Problem 1<\/b><\/span><\/h3><ol><li><b> If arcs AXB and CYD of a circle are congruent, find the ratio of AB and CD.<\/b><\/li><\/ol><h5><span style=\"color: #800000;\"><b>Solution:<\/b><\/span><\/h5><p><span style=\"font-weight: 400;\">Given that arcs AXB and CYD of a circle are congruent,<\/span><\/p><p><span style=\"font-weight: 400;\">i.e. arc AXB \u2245 arc CYD.<\/span><\/p><p><a href=\"https:\/\/mp.moonpreneur.com\/math-corner\/wp-content\/uploads\/2026\/02\/circle_and_tangent_problems_3.jpeg\"><img decoding=\"async\" loading=\"lazy\" class=\"alignnone wp-image-37916\" src=\"https:\/\/mp.moonpreneur.com\/math-corner\/wp-content\/uploads\/2026\/02\/circle_and_tangent_problems_3.jpeg\" alt=\"\" width=\"500\" height=\"374\" \/><\/a><\/p><p><span style=\"font-weight: 400;\">We know that if two arcs of a circle are congruent, their corresponding chords are also equal.<\/span><\/p><p><span style=\"font-weight: 400;\">i.e. chord AB = chord CD<\/span><\/p><p><span style=\"font-weight: 400;\">Thus, AB\/CD = 1<\/span><\/p><p><span style=\"font-weight: 400;\">AB\/CD = 1\/1<\/span><\/p><p><span style=\"font-weight: 400;\">AB : CD = 1 : 1<\/span><\/p><h3><span style=\"color: #008000;\"><b>Problem 2<\/b><\/span><\/h3><p><span style=\"font-weight: 400;\">A circular park of a radius of 20 m is situated in a colony. Three boys Ankur, Syed and David are sitting at an equal distance on the boundary, each having a toy telephone in his hands to talk to each other. Find the length of the string of each phone.<\/span><\/p><h4><span style=\"font-weight: 400; color: #800000;\"><strong>Solution:<\/strong> <\/span><\/h4><p><span style=\"font-weight: 400;\"><a href=\"https:\/\/mp.moonpreneur.com\/math-corner\/wp-content\/uploads\/2026\/02\/circle_and_tangent_problems_2.jpeg\"><img decoding=\"async\" loading=\"lazy\" class=\"alignnone wp-image-37917\" src=\"https:\/\/mp.moonpreneur.com\/math-corner\/wp-content\/uploads\/2026\/02\/circle_and_tangent_problems_2.jpeg\" alt=\"\" width=\"500\" height=\"376\" \/><\/a><\/span><\/p><p><span style=\"font-weight: 400;\">Let A, B, and C be the positions of Ankur, Syed and David, respectively.<\/span><\/p><p><span style=\"font-weight: 400;\">As they are sitting at equal distances, the triangle is equilateral<\/span><\/p><p><span style=\"font-weight: 400;\">AD \u22a5 BC is drawn.<\/span><\/p><p><span style=\"font-weight: 400;\">AD is the median of \u0394ABC, and it passes through the centre O.<\/span><\/p><p><span style=\"font-weight: 400;\">O is the centroid of the \u0394ABC. OA is the radius of the triangle.<\/span><\/p><p><span style=\"font-weight: 400;\">OA = 2\/3 AD<\/span><\/p><p><span style=\"font-weight: 400;\">Let a m be the side of the triangle.<\/span><\/p><p><span style=\"font-weight: 400;\">So, BD = a\/2 m<\/span><\/p><p><span style=\"font-weight: 400;\">In \u0394ABD,<\/span><\/p><p><span style=\"font-weight: 400;\">By Pythagoras theorem,<\/span><\/p><p style=\"text-align: center;\">AB\u00b2 = BD\u00b2 + AD\u00b2<\/p><p style=\"text-align: center;\">AD\u00b2 = AB\u00b2 &#8211; BD\u00b2<\/p><p style=\"text-align: center;\">AD\u00b2 = a\u00b2 &#8211; (a\/2)\u00b2<\/p><p style=\"text-align: center;\">AD\u00b2 = 3a\u00b2\/4<\/p><p style=\"text-align: center;\">AD = \u221a3a\/2<\/p><p style=\"text-align: center;\">OA = 2\/3 AD<\/p><p style=\"text-align: center;\">20 m = (2\/3) \u00d7 (\u221a3a\/2)<\/p><p style=\"text-align: center;\">a = 20\u221a3 m<\/p><h3><span style=\"color: #008000;\"><b>Problem 3<\/b><\/span><\/h3><p>Angle A is circumscribed about circle O.<\/p><p><strong>What is the measure of \u2220A?<\/strong><\/p><p><a href=\"https:\/\/mp.moonpreneur.com\/math-corner\/wp-content\/uploads\/2026\/02\/circle_and_tangent_problems_1.jpeg\"><img decoding=\"async\" loading=\"lazy\" class=\"alignnone wp-image-37918\" src=\"https:\/\/mp.moonpreneur.com\/math-corner\/wp-content\/uploads\/2026\/02\/circle_and_tangent_problems_1.jpeg\" alt=\"\" width=\"500\" height=\"302\" \/><\/a><\/p><h4><span style=\"color: #800000;\"><strong>Solution:<\/strong><\/span><\/h4><p><strong>1. Find the Central Angle<\/strong> (COB)<\/p><p>The Inscribed Angle Theorem states that the measure of the central angle is twice the measure of the inscribed angle that subtends the same arc.<\/p><ul><li>The inscribed angle is\u00a0 \u2220CDB = 65\u00ba<\/li><li>Therefore, the central angle \u2220COB = 2 \u00d7 65\u00ba\u00a0 = 130\u00ba<\/li><\/ul><p><strong>2. Identify the Properties of the Tangents<\/strong><\/p><p>The problem states that angle A is circumscribed about circle O. This means lines AC and AB are tangent to the circle at points C and B.<\/p><ul><li>A radius is always perpendicular to a tangent line at the point of tangency.<\/li><li>So,\u00a0 \u2220ACO = 90\u00ba and \u2220ABO = 90\u00ba<\/li><\/ul><p><strong>3. Solve the Quadrilateral ACOB <\/strong><\/p><p>The sum of the interior angles of any quadrilateral is 360\u00b0. In the quadrilateral ACOB:<\/p><p>\u2220A + \u2220ACO + \u2220COB +\u00a0 \u2220ABO = 360\u00ba<\/p><p>Substitute the values we know:<\/p><p style=\"text-align: center;\">\u2220A + 90\u00ba + 130\u00ba + 90\u00ba = 360\u00ba<\/p><p style=\"text-align: center;\">\u2220A + 310\u00ba = 360\u00ba<\/p><p style=\"text-align: center;\">\u2220A = 360\u00ba &#8211; 310\u00ba<\/p><p style=\"text-align: center;\"><strong>\u2220A = 50\u00ba<\/strong><\/p><h5>\u00a0<\/h5><h5>\u00a0<\/h5>\t\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t<section class=\"has_eae_slider elementor-section elementor-top-section elementor-element elementor-element-000fa3b elementor-section-boxed elementor-section-height-default elementor-section-height-default\" data-id=\"000fa3b\" data-element_type=\"section\">\n\t\t\t\t\t\t<div class=\"elementor-container elementor-column-gap-default\">\n\t\t\t\t\t\t\t<div class=\"elementor-row\">\n\t\t\t\t\t<div class=\"has_eae_slider elementor-column elementor-col-100 elementor-top-column elementor-element elementor-element-1901152\" data-id=\"1901152\" data-element_type=\"column\">\n\t\t\t<div class=\"elementor-column-wrap elementor-element-populated\">\n\t\t\t\t\t\t\t<div class=\"elementor-widget-wrap\">\n\t\t\t\t\t\t<div class=\"elementor-element elementor-element-efb1fbf elementor-widget elementor-widget-text-editor\" data-id=\"efb1fbf\" data-element_type=\"widget\" data-widget_type=\"text-editor.default\">\n\t\t\t\t<div class=\"elementor-widget-container\">\n\t\t\t\t\t\t\t\t<div class=\"elementor-text-editor elementor-clearfix\">\n\t\t\t\t<h4><span style=\"color: #008000;\"><b>Why This Matters for Your Score<\/b><\/span><\/h4><p><span style=\"font-weight: 400;\">Geometry might feel like a small part of the SAT, but these circle problems are &#8220;points on the table.&#8221; Once you recognize the pattern\u2014tangent meets radius, 90-degree angle, right triangle\u2014the math becomes much simpler.<\/span><\/p><p><b><span style=\"color: #008000;\">Pro-tip<\/span>:<\/b><span style=\"font-weight: 400;\"> If you find yourself stuck, look for <\/span><b>concentric circles<\/b><span style=\"font-weight: 400;\"> (circles inside circles). Often, a chord for the big circle is actually a tangent for the small one, which gives you that secret 90-degree angle you need to start solving.<\/span><\/p><p><span style=\"font-weight: 400;\">Keep practicing these &#8220;point of contact&#8221; problems, and soon you&#8217;ll be seeing the hidden triangles in every circle you meet!<\/span><\/p><p><span style=\"font-weight: 400;\">Want to excite your child about math and sharpen their math skills? Moonpreneur&#8217;s online math curriculum is unique as it helps children understand math skills through hands-on lessons, assists them in building real-life applications, and excites them to learn math.\u00a0<\/span><\/p><p><span style=\"font-weight: 400;\">You can opt for our <\/span><a href=\"https:\/\/moonpreneur.com\/innovator-program\/advanced-math\/\"><span style=\"font-weight: 400;\">Advanced Math<\/span><\/a><span style=\"font-weight: 400;\"> or Vedic Math+Mental Math courses. Our <\/span><a href=\"https:\/\/mp.moonpreneur.com\/math-quiz-for-kids\/\"><span style=\"font-weight: 400;\">Math Quiz<\/span><\/a><span style=\"font-weight: 400;\"> for grades 3rd, 4th, 5th, and 6th helps in further exciting and engaging in mathematics with hands-on lessons.<\/span><\/p><p><b>Recommended Reading:<\/b><\/p><ol><li style=\"font-weight: 400;\" aria-level=\"1\"><a href=\"https:\/\/mp.moonpreneur.com\/math-corner\/exponential-equations-using-recursion-and-algebraic\/\"><span style=\"font-weight: 400;\">Solving Exponential Equations Using Recursion: A Step-by-Step Guide<\/span><\/a><\/li><li style=\"font-weight: 400;\" aria-level=\"1\"><a href=\"https:\/\/mp.moonpreneur.com\/math-corner\/linear-equations-different-solutions\/\"><span style=\"font-weight: 400;\">Linear Equation &#8211; One Solution, No Solution and Many Solutions<\/span><\/a><\/li><li style=\"font-weight: 400;\" aria-level=\"1\"><a href=\"https:\/\/mp.moonpreneur.com\/math-corner\/geometry-problem\/\"><span style=\"font-weight: 400;\">Interesting Geometry Problem to Solve For Kids<\/span><\/a><\/li><li style=\"font-weight: 400;\" aria-level=\"1\"><p><a href=\"https:\/\/mp.moonpreneur.com\/math-corner\/sat-quadratics-tricks\/\"><span style=\"font-weight: 400;\">The Ultimate Guide to Solving SAT Quadratics in Seconds<\/span><\/a><\/p><\/li><li style=\"font-weight: 400;\" aria-level=\"1\"><p><a href=\"https:\/\/mp.moonpreneur.com\/math-corner\/derive-quadratic-formula\/\"><span style=\"font-weight: 400;\">How to Derive and Use the Quadratic Formula (With Examples)<\/span><\/a><\/p><\/li><li style=\"font-weight: 400;\" aria-level=\"1\"><a href=\"https:\/\/mp.moonpreneur.com\/math-corner\/sherman-morrison-woodbury-identity\/\"><span style=\"font-weight: 400;\">Application &amp; Proof\u00a0 of the Sherman-Morrison-Woodbury Identity<\/span><\/a><\/li><li style=\"font-weight: 400;\" aria-level=\"1\"><a href=\"https:\/\/mp.moonpreneur.com\/math-corner\/geometric-problem-unsolved-by-ai\/\"><span style=\"font-weight: 400;\">The Geometry Problem That Still Defeats ChatGPT, Gemini, and Grok<\/span><\/a><\/li><\/ol>\t\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t<section class=\"has_eae_slider elementor-section elementor-top-section elementor-element elementor-element-4dd7e1e elementor-section-boxed elementor-section-height-default elementor-section-height-default\" data-id=\"4dd7e1e\" data-element_type=\"section\">\n\t\t\t\t\t\t<div class=\"elementor-container elementor-column-gap-default\">\n\t\t\t\t\t\t\t<div class=\"elementor-row\">\n\t\t\t\t\t<div class=\"has_eae_slider elementor-column elementor-col-100 elementor-top-column elementor-element elementor-element-70c82a2\" data-id=\"70c82a2\" data-element_type=\"column\">\n\t\t\t<div class=\"elementor-column-wrap elementor-element-populated\">\n\t\t\t\t\t\t\t<div class=\"elementor-widget-wrap\">\n\t\t\t\t\t\t<div class=\"elementor-element elementor-element-eb6ea7a elementor-widget elementor-widget-text-editor\" data-id=\"eb6ea7a\" data-element_type=\"widget\" data-widget_type=\"text-editor.default\">\n\t\t\t\t<div class=\"elementor-widget-container\">\n\t\t\t\t\t\t\t\t<div class=\"elementor-text-editor elementor-clearfix\">\n\t\t\t\t<h3><strong>FAQs on\u00a0 Geometry Problems<\/strong><\/h3>\t\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t<section class=\"has_eae_slider elementor-section elementor-top-section elementor-element elementor-element-44c18e9 elementor-section-boxed elementor-section-height-default elementor-section-height-default\" data-id=\"44c18e9\" data-element_type=\"section\">\n\t\t\t\t\t\t<div class=\"elementor-container elementor-column-gap-default\">\n\t\t\t\t\t\t\t<div class=\"elementor-row\">\n\t\t\t\t\t<div class=\"has_eae_slider elementor-column elementor-col-100 elementor-top-column elementor-element elementor-element-24f8067\" data-id=\"24f8067\" data-element_type=\"column\">\n\t\t\t<div class=\"elementor-column-wrap elementor-element-populated\">\n\t\t\t\t\t\t\t<div class=\"elementor-widget-wrap\">\n\t\t\t\t\t\t<div class=\"elementor-element elementor-element-b4d1436 elementor-widget elementor-widget-elementskit-faq\" data-id=\"b4d1436\" data-element_type=\"widget\" data-widget_type=\"elementskit-faq.default\">\n\t\t\t\t<div class=\"elementor-widget-container\">\n\t\t\t<div class=\"ekit-wid-con\">\n                <div class=\"elementskit-single-faq elementor-repeater-item-95228b5\">\n            <div class=\"elementskit-faq-header\">\n                <h2 class=\"elementskit-faq-title\">Q1. What is the tangent to a circle?<\/h2>\n            <\/div>\n            <div class=\"elementskit-faq-body\">\n                Ans. A tangent to a circle is a line that touches the circle at exactly one point, called the point of tangency. It is perpendicular to the radius drawn to that point and does not intersect the circle at any other point.\n            <\/div>\n        <\/div>\n                <div class=\"elementskit-single-faq elementor-repeater-item-62a8206\">\n            <div class=\"elementskit-faq-header\">\n                <h2 class=\"elementskit-faq-title\">Q2. What is the tangent-radius theorem?<\/h2>\n            <\/div>\n            <div class=\"elementskit-faq-body\">\n                Ans.The tangent-radius theorem states that a tangent line to a circle is always perpendicular to the radius drawn to the point of tangency. This means that the angle between the tangent line and the radius at the point of tangency is always 90 degrees.\n            <\/div>\n        <\/div>\n                <div class=\"elementskit-single-faq elementor-repeater-item-80e5d28\">\n            <div class=\"elementskit-faq-header\">\n                <h2 class=\"elementskit-faq-title\">Q3. How do you find the equation of a tangent line to a circle?<\/h2>\n            <\/div>\n            <div class=\"elementskit-faq-body\">\n                Ans. To find the equation of a tangent line to a circle, you need to know the circle's equation and the point of tangency. The general steps are: Find the slope of the radius at the point of tangency, calculate the negative reciprocal of this slope to get the slope of the tangent line, Use the point-slope form of a line equation with the point of tangency and the calculated slope.\n            <\/div>\n        <\/div>\n                <div class=\"elementskit-single-faq elementor-repeater-item-3b695cd\">\n            <div class=\"elementskit-faq-header\">\n                <h2 class=\"elementskit-faq-title\">Q4. What is the two tangent theorem?<\/h2>\n            <\/div>\n            <div class=\"elementskit-faq-body\">\n                Ans. The two tangent theorem states that if two tangent lines are drawn to a circle from an external point, the lengths of these tangent segments are equal. This theorem is useful in solving problems involving tangents and external points.\n            <\/div>\n        <\/div>\n        \n    <\/div>\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t\t\t\t\t\t<\/div>\n\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t","protected":false},"excerpt":{"rendered":"<p>If you\u2019ve ever looked at a geometry problem involving circles and felt like you were going in&#8230; well, circles&#8230; You aren&#8217;t alone. For many SAT students, the geometry section can feel like a collection of abstract rules, but when it comes to tangents, there are actually some very reliable &#8220;cheat codes&#8221; you can use to [&hellip;]<\/p>\n","protected":false},"author":116,"featured_media":38068,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"inline_featured_image":false},"categories":[979,1034,986],"tags":[],"acf":[],"_links":{"self":[{"href":"https:\/\/mp.moonpreneur.com\/math-corner\/wp-json\/wp\/v2\/posts\/37914"}],"collection":[{"href":"https:\/\/mp.moonpreneur.com\/math-corner\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/mp.moonpreneur.com\/math-corner\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/mp.moonpreneur.com\/math-corner\/wp-json\/wp\/v2\/users\/116"}],"replies":[{"embeddable":true,"href":"https:\/\/mp.moonpreneur.com\/math-corner\/wp-json\/wp\/v2\/comments?post=37914"}],"version-history":[{"count":12,"href":"https:\/\/mp.moonpreneur.com\/math-corner\/wp-json\/wp\/v2\/posts\/37914\/revisions"}],"predecessor-version":[{"id":38072,"href":"https:\/\/mp.moonpreneur.com\/math-corner\/wp-json\/wp\/v2\/posts\/37914\/revisions\/38072"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/mp.moonpreneur.com\/math-corner\/wp-json\/wp\/v2\/media\/38068"}],"wp:attachment":[{"href":"https:\/\/mp.moonpreneur.com\/math-corner\/wp-json\/wp\/v2\/media?parent=37914"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mp.moonpreneur.com\/math-corner\/wp-json\/wp\/v2\/categories?post=37914"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mp.moonpreneur.com\/math-corner\/wp-json\/wp\/v2\/tags?post=37914"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}