{"id":63959,"date":"2023-09-05T15:04:00","date_gmt":"2023-09-05T09:34:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=63959"},"modified":"2023-11-02T10:50:07","modified_gmt":"2023-11-02T05:20:07","slug":"rd-sharma-solutions-class-11-maths-chapter-24-the-circle","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-24-the-circle\/","title":{"rendered":"RD Sharma Solutions Class 11 Maths Chapter 24 &#8211; The Circle (Updated For 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-121070\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-11-Maths-Chapter-24-The-Circle.jpg\" alt=\"RD Sharma Solutions Class 11 Maths Chapter 24 \" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-11-Maths-Chapter-24-The-Circle.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-11-Maths-Chapter-24-The-Circle-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Solutions Class 11 Maths Chapter 24<\/strong> &#8211; The Circle: RD Sharma Solutions Class 11 Maths Chapter 24 The Circle assists the students in getting to know about the exam pattern so as to achieve higher marks in exams. The solutions are developed with shortcut techniques that make learning easy. The students can attain good marks in the final exams by practicing these solutions regularly from <a href=\"https:\/\/www.kopykitab.com\/blog\/cbse-class-11-maths-rd-sharma-solutions\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 11 Maths<\/a>.<\/p>\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_47_1 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"ez-toc-toggle-icon-1\"><label for=\"item-69e73e6d655ae\" aria-label=\"Table of Content\"><span style=\"display: flex;align-items: center;width: 35px;height: 30px;justify-content: center;direction:ltr;\"><svg style=\"fill: #000000;color:#000000\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #000000;color:#000000\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/label><input  type=\"checkbox\" id=\"item-69e73e6d655ae\"><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 eztoc-visibility-hide-by-default' ><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-24-the-circle\/#download-rd-sharma-solutions-class-11-maths-chapter-24-%e2%80%93-the-circle-pdf\" title=\"Download RD Sharma Solutions Class 11 Maths Chapter 24 &#8211; The Circle PDF\">Download RD Sharma Solutions Class 11 Maths Chapter 24 &#8211; The Circle PDF<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-24-the-circle\/#exercise-wise-rd-sharma-solutions-class-11-maths-chapter-24-pdf\" title=\"Exercise-wise RD Sharma Solutions Class 11 Maths Chapter 24 PDF\">Exercise-wise RD Sharma Solutions Class 11 Maths Chapter 24 PDF<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-24-the-circle\/#detailed-exercise-wise-explanation-with-a-listing-of-important-topics-in-the-exercise\" title=\"Detailed Exercise-wise Explanation with a Listing of Important Topics in the Exercise\">Detailed Exercise-wise Explanation with a Listing of Important Topics in the Exercise<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-24-the-circle\/#rd-sharma-class-11-chapter-24-exercise-24a\" title=\"RD Sharma Class 11 Chapter 24 Exercise 24a\">RD Sharma Class 11 Chapter 24 Exercise 24a<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-24-the-circle\/#rd-sharma-class-11-chapter-24-exercise-24b\" title=\"RD Sharma Class 11 Chapter 24 Exercise 24b\">RD Sharma Class 11 Chapter 24 Exercise 24b<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-24-the-circle\/#rd-sharma-class-11-chapter-24-exercise-24c\" title=\"RD Sharma Class 11 Chapter 24 Exercise 24c\">RD Sharma Class 11 Chapter 24 Exercise 24c<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-7\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-24-the-circle\/#access-rd-sharma-solutions-class-11-maths-chapter-24\" title=\"Access RD Sharma Solutions Class 11 Maths Chapter 24\">Access RD Sharma Solutions Class 11 Maths Chapter 24<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-8\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-24-the-circle\/#important-topics-class-11-maths-chapter-24-%e2%80%93-the-circle\" title=\"Important Topics Class 11 Maths Chapter 24 &#8211; The Circle\">Important Topics Class 11 Maths Chapter 24 &#8211; The Circle<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-9\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-24-the-circle\/#rd-sharma-solutions-class-11-maths-chapter-24-%e2%80%93-the-circle\" title=\"RD Sharma Solutions Class 11 Maths Chapter 24 &#8211; The Circle\">RD Sharma Solutions Class 11 Maths Chapter 24 &#8211; The Circle<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-10\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-24-the-circle\/#faqs-on-rd-sharma-solutions-class-11-maths-chapter-24\" title=\"FAQs on RD Sharma Solutions Class 11 Maths Chapter 24\">FAQs on RD Sharma Solutions Class 11 Maths Chapter 24<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-11\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-24-the-circle\/#can-i-access-the-rd-sharma-solutions-class-11-maths-chapter-24-pdf-offline\" title=\"Can I access the RD Sharma Solutions Class 11 Maths Chapter 24\u00a0PDF offline?\">Can I access the RD Sharma Solutions Class 11 Maths Chapter 24\u00a0PDF offline?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-12\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-24-the-circle\/#how-much-does-it-cost-to-download-the-pdf-of-rd-sharma-solutions-class-11-maths-chapter-24\" title=\"How much does it cost to download the PDF of RD Sharma Solutions Class 11 Maths Chapter 24?\">How much does it cost to download the PDF of RD Sharma Solutions Class 11 Maths Chapter 24?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-13\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-24-the-circle\/#from-where-can-i-download-the-pdf-of-rd-sharma-solutions-class-11-maths-chapter-24\" title=\"From where can I download the PDF of RD Sharma Solutions Class 11 Maths Chapter 24?\">From where can I download the PDF of RD Sharma Solutions Class 11 Maths Chapter 24?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-solutions-class-11-maths-chapter-24-%e2%80%93-the-circle-pdf\"><\/span>Download RD Sharma Solutions Class 11 Maths Chapter 24 &#8211; The Circle PDF<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-11-Maths-Chapter-24.pdf\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 11 Maths Chapter 24<\/a><\/p>\n<div id=\"example1\" style=\"text-align: justify;\">\u00a0<\/div>\n<p style=\"text-align: justify;\"><style>\n.pdfobject-container { height: 500px;}<br \/>\n.pdfobject { border: 1px solid #666; }<br \/>\n<\/style><\/p>\n<p style=\"text-align: justify;\"><script src=\"https:\/\/www.kopykitab.com\/_utility\/js\/pdfobject.min.js\"><\/script><br \/><script>PDFObject.embed(\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-11-Maths-Chapter-24.pdf\",\"#example1\");<\/script><\/p>\n<h2><span class=\"ez-toc-section\" id=\"exercise-wise-rd-sharma-solutions-class-11-maths-chapter-24-pdf\"><\/span>Exercise-wise RD Sharma Solutions Class 11 Maths Chapter 24 PDF<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<table style=\"border-collapse: collapse; width: 100%;\">\n<tbody>\n<tr>\n<td style=\"width: 100%;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-24-exercise-24-1\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Class 11 Chapter 24 Exercise 24a<\/a><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 100%;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-24-exercise-24-2\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Class 11 Chapter 24 Exercise 24b<\/a><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 100%;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-24-exercise-24-3\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Class 11 Chapter 24 Exercise 24c<\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h2><span class=\"ez-toc-section\" id=\"detailed-exercise-wise-explanation-with-a-listing-of-important-topics-in-the-exercise\"><\/span>Detailed Exercise-wise Explanation with a Listing of Important Topics in the Exercise<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>RD Sharma Solutions Class 11 Maths Chapter 24 are solved by experienced maths faculty with easy methods. These solutions include well-researched information about each exercise so that the students can easily understand all concepts.<\/p>\n<p>These solutions enable the students to get confident about their answers by practising RD Sharma&#8217;s class 11 solutions Maths Chapter 24 The Circle on a regular basis. The students can refer to exercise-wise solutions provided with short keynotes that can boost their exam preparations.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-class-11-chapter-24-exercise-24a\"><\/span><strong>RD Sharma Class 11 Chapter 24 Exercise 24a<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>This exercise includes finding the equation of a circle when its center &amp; radius are known &amp; finding the center and radius of a given circle. RD Sharma class 11 chapter 24 exercise 24a assists the students to get excellent marks in the exams.<\/p>\n<p>The solutions are prepared by the maths expert faculty in a step-by-step manner to assist the students in understanding the topics clearly. The students must practice exercise-wise problems regularly in order to attain better marks in the exams.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-class-11-chapter-24-exercise-24b\"><\/span>RD Sharma Class 11 Chapter 24 Exercise 24b<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>This exercise includes problems related to the topics of the general equation of a circle. The students can strengthen their knowledge &amp; build a strong command over the topics of this chapter with the help of these solutions.<\/p>\n<p>The Mathematics experts have solved RD Sharma Class 11 Chapter 24 Exercise 24b in an easy &amp; understandable language.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-class-11-chapter-24-exercise-24c\"><\/span>RD Sharma Class 11 Chapter 24 Exercise 24c<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>This exercise includes problems based on the topics of the diameter form of a circle. The students can refer to this exercise solution as a reference guide to enhance their conceptual knowledge &amp; understand the different methods to solve the problems.<\/p>\n<p>RD Sharma class 11 chapter 24 exercise 24c is created by the Maths expert faculty to improve students\u2019 confidence to get good marks.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"access-rd-sharma-solutions-class-11-maths-chapter-24\"><\/span>Access RD Sharma Solutions Class 11 Maths Chapter 24<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><strong>EXERCISE 24.1 PAGE NO: 24.21<\/strong><\/p>\n<p><strong>1. Find the equation of the circle with:<br \/>(i) Centre (-2, 3) and radius 4.<\/strong><\/p>\n<p><strong>(ii) Centre (a, b) and radius<img title=\"RD Sharma Solutions for Class 11 Maths Chapter 24 \u2013 The Circle\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-24-the-circle-image-1.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 24 \u2013 The Circle - image 1\" \/>.<\/strong><\/p>\n<p><strong>(iii) Centre (0, \u2013 1) and radius 1.<\/strong><\/p>\n<p><strong>(iv) Centre (a cos \u03b1, a sin \u03b1) and radius a.<\/strong><\/p>\n<p><strong>(v) Centre (a, a) and radius \u221a2\u00a0a.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong>(i)\u00a0<\/strong>Centre (-2, 3) and radius 4.<\/p>\n<p>Given:<\/p>\n<p>The radius is 4, and the center (-2, 3)<\/p>\n<p>By using the formula,<\/p>\n<p>The equation of the circle with center (p, q) and radius \u2018r\u2019 is (x \u2013 p)<sup>2<\/sup>\u00a0+ (y \u2013 q)<sup>2<\/sup>\u00a0= r<sup>2<\/sup><\/p>\n<p>Where, p = -2, q = 3, r = 4<\/p>\n<p>Now, by substituting the values in the above equation, we get<\/p>\n<p>(x \u2013 p)<sup>2<\/sup>\u00a0+ (y \u2013 q)<sup>2<\/sup>\u00a0= r<sup>2<\/sup><\/p>\n<p>(x \u2013 (-2))<sup>2<\/sup>\u00a0+ (y \u2013 3)<sup>2<\/sup>\u00a0= 4<sup>2<\/sup><\/p>\n<p>(x + 2)<sup>2<\/sup>\u00a0+ (y \u2013 3)<sup>2<\/sup>\u00a0= 16<\/p>\n<p>x<sup>2<\/sup>\u00a0+ 4x + 4 + y<sup>2<\/sup>\u00a0\u2013 6y + 9 = 16<\/p>\n<p>x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0+ 4x \u2013 6y \u2013 3 = 0<\/p>\n<p><em>\u2234 The equation of the circle is x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0+ 4x \u2013 6y \u2013 3 = 0<\/em><\/p>\n<p><strong>(ii)\u00a0<\/strong>Centre (a, b) and radius<br \/><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 24 \u2013 The Circle\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-24-the-circle-image-2.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 24 \u2013 The Circle - image 2\" \/>.<\/p>\n<p>Given:<\/p>\n<p>The radius is<br \/><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 24 \u2013 The Circle\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-24-the-circle-image-3.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 24 \u2013 The Circle - image 3\" \/>and the centre (a, b)<\/p>\n<p>By using the formula,<\/p>\n<p>The equation of the circle with center (p, q) and radius \u2018r\u2019 is (x \u2013 p)<sup>2<\/sup>\u00a0+ (y \u2013 q)<sup>2<\/sup>\u00a0= r<sup>2<\/sup><\/p>\n<p>Where, p = a, q = b, r =<br \/><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 24 \u2013 The Circle\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-24-the-circle-image-4.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 24 \u2013 The Circle - image 4\" \/><\/p>\n<p>Now, by substituting the values in the above equation, we get<\/p>\n<p>(x \u2013 p)<sup>2<\/sup>\u00a0+ (y \u2013 q)<sup>2<\/sup>\u00a0= r<sup>2<\/sup><\/p>\n<p>(x \u2013 a)<sup>2<\/sup>\u00a0+ (y \u2013 b)<sup>2<\/sup>\u00a0=<br \/><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 24 \u2013 The Circle\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-24-the-circle-image-5.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 24 \u2013 The Circle - image 5\" \/><\/p>\n<p>x<sup>2<\/sup>\u00a0\u2013 2ax + a<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0\u2013 2by + b<sup>2<\/sup>\u00a0= a<sup>2<\/sup>\u00a0+ b<sup>2<\/sup><\/p>\n<p>x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0\u2013 2ax \u2013 2by = 0<\/p>\n<p><em>\u2234 The equation of the circle is x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0\u2013 2ax \u2013 2by = 0<\/em><\/p>\n<p><strong>(iii)\u00a0<\/strong>Centre (0, -1) and radius 1.<\/p>\n<p>Given:<\/p>\n<p>The radius is 1, and the center (0, -1)<\/p>\n<p>By using the formula,<\/p>\n<p>The equation of the circle with center (p, q) and radius \u2018r\u2019 is (x \u2013 p)<sup>2<\/sup>\u00a0+ (y \u2013 q)<sup>2<\/sup>\u00a0= r<sup>2<\/sup><\/p>\n<p>Where, p = 0, q = -1, r = 1<\/p>\n<p>Now, by substituting the values in the above equation, we get<\/p>\n<p>(x \u2013 p)<sup>2<\/sup>\u00a0+ (y \u2013 q)<sup>2<\/sup>\u00a0= r<sup>2<\/sup><\/p>\n<p>(x \u2013 0)<sup>2<\/sup>\u00a0+ (y \u2013 (-1))<sup>2<\/sup>\u00a0= 1<sup>2<\/sup><\/p>\n<p>(x \u2013 0)<sup>2<\/sup>\u00a0+ (y + 1)<sup>2<\/sup>\u00a0= 1<\/p>\n<p>x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0+ 2y + 1 = 1<\/p>\n<p>x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0+ 2y = 0<\/p>\n<p><em>\u2234 The equation of the circle is x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0+ 2y = 0.<\/em><\/p>\n<p><strong>(iv)\u00a0<\/strong>Centre (a cos \u03b1, a sin \u03b1) and radius a.<\/p>\n<p>Given:<\/p>\n<p>The radius is \u2018a\u2019, and the center (a cos \u03b1, a sin \u03b1)<\/p>\n<p>By using the formula,<\/p>\n<p>The equation of the circle with center (p, q) and radius \u2018r\u2019 is (x \u2013 p)<sup>2<\/sup>\u00a0+ (y \u2013 q)<sup>2<\/sup>\u00a0= r<sup>2<\/sup><\/p>\n<p>Where, p = a cos \u03b1, q = a sin \u03b1, r = a<\/p>\n<p>Now, by substituting the values in the above equation, we get<\/p>\n<p>(x \u2013 p)<sup>2<\/sup>\u00a0+ (y \u2013 q)<sup>2<\/sup>\u00a0= r<sup>2<\/sup><\/p>\n<p>(x \u2013 a cos\u03b1)<sup>2<\/sup>\u00a0+ (y \u2013 a sin\u03b1)<sup>2<\/sup>\u00a0= a<sup>2<\/sup><\/p>\n<p>x<sup>2<\/sup>\u00a0\u2013 (2acos\u03b1)x + a<sup>2<\/sup>cos<sup>2<\/sup>\u03b1 + y<sup>2<\/sup>\u00a0\u2013 (2asin\u03b1)y + a<sup>2<\/sup>sin<sup>2<\/sup>\u03b1 = a<sup>2<\/sup><\/p>\n<p>We know that sin<sup>2<\/sup>\u03b8 + cos<sup>2<\/sup>\u03b8 = 1<\/p>\n<p>So,<\/p>\n<p>x<sup>2<\/sup>\u00a0\u2013 (2acos\u03b1)x + y<sup>2<\/sup>\u00a0\u2013 2asin\u03b1y + a<sup>2<\/sup>\u00a0= a<sup>2<\/sup><\/p>\n<p>x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0\u2013 (2acos\u03b1)x \u2013 (2asin\u03b1)y\u00a0= 0<\/p>\n<p><em>\u2234 The equation of the circle is x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0\u2013 (2acos\u03b1) x \u2013 (2asin\u03b1) y = 0.<\/em><\/p>\n<p><strong>(v)\u00a0<\/strong>Centre (a, a) and radius \u221a2\u00a0a.<\/p>\n<p>Given:<\/p>\n<p>The radius is \u221a2 a, and the center (a, a)<\/p>\n<p>By using the formula,<\/p>\n<p>The equation of the circle with center (p, q) and radius \u2018r\u2019 is (x \u2013 p)<sup>2<\/sup>\u00a0+ (y \u2013 q)<sup>2<\/sup>\u00a0= r<sup>2<\/sup><\/p>\n<p>Where, p = a, q = a, r = \u221a2\u00a0a<\/p>\n<p>Now, by substituting the values in the above equation, we get<\/p>\n<p>(x \u2013 p)<sup>2<\/sup>\u00a0+ (y \u2013 q)<sup>2<\/sup>\u00a0= r<sup>2<\/sup><\/p>\n<p>(x \u2013 a)<sup>2<\/sup>\u00a0+ (y \u2013 a)<sup>2<\/sup>\u00a0= (\u221a2\u00a0a)<sup>2<\/sup><\/p>\n<p>x<sup>2<\/sup>\u00a0\u2013 2ax + a<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0\u2013 2ay + a<sup>2<\/sup>\u00a0= 2a<sup>2<\/sup><\/p>\n<p>x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0\u2013 2ax \u2013 2ay = 0<\/p>\n<p><em>\u2234 The equation of the circle is x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0\u2013 2ax \u2013 2ay = 0.<\/em><\/p>\n<p><strong>2. Find the center and radius of each of the following circles:<br \/>(i) (x \u2013 1)<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0= 4<\/strong><\/p>\n<p><strong>(ii) (x + 5)<sup>2<\/sup>\u00a0+ (y + 1)<sup>2<\/sup>\u00a0= 9<\/strong><\/p>\n<p><strong>(iii) x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0\u2013 4x + 6y = 5<\/strong><\/p>\n<p><strong>(iv) x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0\u2013 x + 2y \u2013 3 = 0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong>(i)\u00a0<\/strong>(x \u2013 1)<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0= 4<\/p>\n<p>Given:<\/p>\n<p>The equation (x \u2013 1)<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0= 4<\/p>\n<p>We need to find the center and the radius.<\/p>\n<p>By using the standard equation formula,<\/p>\n<p>(x \u2013 a)<sup>2<\/sup>\u00a0+ (y \u2013 b)<sup>2<\/sup>\u00a0= r<sup>2<\/sup>\u00a0\u2026. (1)<\/p>\n<p>Now, let us convert the given circle\u2019s equation into the standard form.<\/p>\n<p>(x \u2013 1)<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0= 4<\/p>\n<p>(x \u2013 1)<sup>2<\/sup>\u00a0+ (y \u2013 0)<sup>2<\/sup>\u00a0= 2<sup>2<\/sup>\u00a0\u2026.. (2)<\/p>\n<p>By comparing equation (2) with (1), we get<\/p>\n<p>Centre = (1, 0) and radius = 2<\/p>\n<p><em>\u2234 The center of the circle is (1, 0), and the radius is 2.<\/em><\/p>\n<p><strong>(ii)\u00a0<\/strong>(x + 5)<sup>2<\/sup>\u00a0+ (y + 1)<sup>2<\/sup>\u00a0= 9<\/p>\n<p>Given:<\/p>\n<p>The equation (x + 5)<sup>2<\/sup>\u00a0+ (y + 1)<sup>2<\/sup>\u00a0= 9<\/p>\n<p>We need to find the center and the radius.<\/p>\n<p>By using the standard equation formula,<\/p>\n<p>(x \u2013 a)<sup>2<\/sup>\u00a0+ (y \u2013 b)<sup>2<\/sup>\u00a0= r<sup>2<\/sup>\u00a0\u2026. (1)<\/p>\n<p>Now let us convert the given circle\u2019s equation into the standard form.<\/p>\n<p>(x + 5)<sup>2<\/sup>\u00a0+ (y + 1)<sup>2<\/sup>\u00a0= 9<\/p>\n<p>(x \u2013 (-5))<sup>2<\/sup>\u00a0+ (y \u2013 ( \u2013 1))<sup>2<\/sup>\u00a0= 3<sup>2<\/sup>\u00a0\u2026. (2)<\/p>\n<p>By comparing equation (2) with (1), we get<\/p>\n<p>Centre = (-5, -1) and radius = 3<\/p>\n<p><em>\u2234 The center of the circle is (-5, -1), and the radius is 3.<\/em><\/p>\n<p><strong>(iii)\u00a0<\/strong>x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0\u2013 4x + 6y = 5<\/p>\n<p>Given:<\/p>\n<p>The equation x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0\u2013 4x + 6y = 5<\/p>\n<p>We need to find the center and the radius.<\/p>\n<p>By using the standard equation formula,<\/p>\n<p>(x \u2013 a)<sup>2<\/sup>\u00a0+ (y \u2013 b)<sup>2<\/sup>\u00a0= r<sup>2<\/sup>\u00a0\u2026. (1)<\/p>\n<p>Now let us convert the given circle\u2019s equation into the standard form.<\/p>\n<p>x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0\u2013 4x + 6y = 5<\/p>\n<p>(x<sup>2<\/sup>\u00a0\u2013 4x + 4) + (y<sup>2<\/sup>\u00a0+ 6y + 9) = 5 + 4 + 9<\/p>\n<p>(x \u2013 2)<sup>2<\/sup>\u00a0+ (y + 3)<sup>2<\/sup>\u00a0= 18<\/p>\n<p>(x \u2013 2)<sup>2<\/sup>\u00a0+ (y \u2013 (-3))<sup>2<\/sup>\u00a0= (3\u221a2)<sup>2<\/sup>\u00a0\u2026 (2)<\/p>\n<p>By comparing equation (2) with (1), we get<\/p>\n<p>Centre = (2, -3) and radius = 3\u221a2<\/p>\n<p><em>\u2234 The center of the circle is (2, -3), and the radius is 3\u221a2.<\/em><\/p>\n<p><strong>(iv)\u00a0<\/strong>x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0\u2013 x + 2y \u2013 3 = 0<\/p>\n<p>Given:<\/p>\n<p>The equation x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0\u2013 x + 2y \u2013 3 = 0<\/p>\n<p>We need to find the center and the radius.<\/p>\n<p>By using the standard equation formula,<\/p>\n<p>(x \u2013 a)<sup>2<\/sup>\u00a0+ (y \u2013 b)<sup>2<\/sup>\u00a0= r<sup>2<\/sup>\u00a0\u2026. (1)<\/p>\n<p>Now let us convert the given circle\u2019s equation into the standard form.<\/p>\n<p>x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0\u2013 x + 2y \u2013 3 = 0<\/p>\n<p>(x<sup>2<\/sup>\u00a0\u2013 x + \u00bc) + (y<sup>2<\/sup>\u00a0+ 2y + 1) \u2013 3 \u2013 \u00bc \u2013 1 = 0<\/p>\n<p>(x \u2013 \u00bd)<sup>2<\/sup>\u00a0+ (y + 1)<sup>2<\/sup>\u00a0= 17\/4 \u2026. (2)<\/p>\n<p>By comparing equation (2) with (1), we get<\/p>\n<p>Centre = (\u00bd, \u2013 1) and radius = \u221a17\/2<\/p>\n<p><em>\u2234 The center of the circle is (\u00bd, -1), and the radius is \u221a17\/2.<\/em><\/p>\n<p><strong>3. Find the equation of the circle whose center is (1, 2) and which passes through the point (4, 6).<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>The Centre is (1, 2) and passes through the point (4, 6).<\/p>\n<p>Where, p = 1, q = 2<\/p>\n<p>We need to find the equation of the circle.<\/p>\n<p>By using the formula,<\/p>\n<p>(x \u2013 p)<sup>2<\/sup>\u00a0+ (y \u2013 q)<sup>2<\/sup>\u00a0= r<sup>2<\/sup><\/p>\n<p>(x \u2013 1)<sup>2<\/sup>\u00a0+ (y \u2013 2)<sup>2<\/sup>\u00a0= r<sup>2<\/sup><\/p>\n<p>It passes through the point (4, 6)<\/p>\n<p>(4 \u2013 1)<sup>2<\/sup>\u00a0+ (6 \u2013 2)<sup>2<\/sup>\u00a0= r<sup>2<\/sup><\/p>\n<p>3<sup>2<\/sup>\u00a0+ 4<sup>2<\/sup>\u00a0= r<sup>2<\/sup><\/p>\n<p>9 + 16 = r<sup>2<\/sup><\/p>\n<p>25 = r<sup>2<\/sup><\/p>\n<p>r = \u221a25<\/p>\n<p>= 5<\/p>\n<p>So r = 5 units<\/p>\n<p>We know that the equation of the circle with center (p, q) and having radius \u2018r\u2019 is given by: (x \u2013 p)<sup>2<\/sup>\u00a0+ (y \u2013 q)<sup>2<\/sup>\u00a0= r<sup>2<\/sup><\/p>\n<p>By substituting the values in the above equation, we get<\/p>\n<p>(x \u2013 1)<sup>2<\/sup>\u00a0+ (y \u2013 2)<sup>2<\/sup>\u00a0= 5<sup>2<\/sup><\/p>\n<p>x<sup>2<\/sup>\u00a0\u2013 2x + 1 + y<sup>2<\/sup>\u00a0\u2013 4y + 4 = 25<\/p>\n<p>x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0\u2013 2x \u2013 4y \u2013 20 = 0.<\/p>\n<p><em>\u2234 The equation of the circle is x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0\u2013 2x \u2013 4y \u2013 20 = 0.<\/em><\/p>\n<h2><span class=\"ez-toc-section\" id=\"important-topics-class-11-maths-chapter-24-%e2%80%93-the-circle\"><\/span>Important Topics Class 11 Maths Chapter 24 &#8211; The Circle<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>RD Sharma Solutions Class 11 Maths Chapter 24 covers some important concepts that are listed below:<\/p>\n<ul>\n<li>Standard equation of a circle.<\/li>\n<li>General equations of a circle.<\/li>\n<li>Diameter form of a circle.<\/li>\n<\/ul>\n<h2><span class=\"ez-toc-section\" id=\"rd-sharma-solutions-class-11-maths-chapter-24-%e2%80%93-the-circle\"><\/span>RD Sharma Solutions Class 11 Maths Chapter 24 &#8211; The Circle<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>RD Sharma Solutions Class 11 Maths Chapter 24 enables the students to learn about the circle &amp; also find the equation of any circle whose center &amp; radius are given. These solutions are provided in the most comprehensible &amp; easy manner. The solutions are solved using shortcut techniques to enable the students to grasp the topics quickly &amp; easily.<\/p>\n<p>The students must go refer to RD Sharma&#8217;s class 11 Solutions Maths chapter 24 The Circle to acquire better marks in final exams. If students can practice exercise-wise solutions on a regular basis, they can definitely attain good marks in the exams. These solutions are accurate and reliable &amp; solved as per the <a href=\"https:\/\/cbse.nic.in\/\" target=\"_blank\" rel=\"noopener noreferrer\">CBSE<\/a> guidelines.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-solutions-class-11-maths-chapter-24\"><\/span>FAQs on RD Sharma Solutions Class 11 Maths Chapter 24<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n<div id=\"rank-math-faq\" class=\"rank-math-block\">\n<div class=\"rank-math-list \">\n<div id=\"faq-question-1629996811750\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"can-i-access-the-rd-sharma-solutions-class-11-maths-chapter-24-pdf-offline\"><\/span>Can I access the RD Sharma Solutions Class 11 Maths Chapter 24\u00a0PDF offline?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Once you have downloaded the PDF online, you can access it offline as well.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1629996850508\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"how-much-does-it-cost-to-download-the-pdf-of-rd-sharma-solutions-class-11-maths-chapter-24\"><\/span>How much does it cost to download the PDF of RD Sharma Solutions Class 11 Maths Chapter 24?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can download it for free.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1629996971491\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"from-where-can-i-download-the-pdf-of-rd-sharma-solutions-class-11-maths-chapter-24\"><\/span>From where can I download the PDF of RD Sharma Solutions Class 11 Maths Chapter 24?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can find the download link from the above blog.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>RD Sharma Solutions Class 11 Maths Chapter 24 &#8211; The Circle: RD Sharma Solutions Class 11 Maths Chapter 24 The Circle assists the students in getting to know about the exam pattern so as to achieve higher marks in exams. The solutions are developed with shortcut techniques that make learning easy. The students can attain &#8230; <a title=\"RD Sharma Solutions Class 11 Maths Chapter 24 &#8211; The Circle (Updated For 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-24-the-circle\/\" aria-label=\"More on RD Sharma Solutions Class 11 Maths Chapter 24 &#8211; The Circle (Updated For 2024)\">Read more<\/a><\/p>\n","protected":false},"author":243,"featured_media":121070,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[2985,73411,73410],"tags":[3428,62517,73334],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/63959"}],"collection":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/users\/243"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=63959"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/63959\/revisions"}],"predecessor-version":[{"id":501133,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/63959\/revisions\/501133"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/121070"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=63959"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=63959"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=63959"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}