{"id":68067,"date":"2023-05-22T09:50:00","date_gmt":"2023-05-22T04:20:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=68067"},"modified":"2023-10-25T15:16:04","modified_gmt":"2023-10-25T09:46:04","slug":"rd-sharma-solutions-class-11-maths-chapter-24-exercise-24-1","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-24-exercise-24-1\/","title":{"rendered":"RD Sharma Class 11 Solutions Chapter 24 Exercise 24.1 (Updated for 2023)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-123795\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Class-11-Solutions-Chapter-24-Exercise-24.1.jpg\" alt=\"RD Sharma Solutions Class 11 Maths Chapter 24 Exercise 24.1\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Class-11-Solutions-Chapter-24-Exercise-24.1.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Class-11-Solutions-Chapter-24-Exercise-24.1-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Solutions Class 11 Maths Chapter 24 Exercise 24.1:\u00a0<\/strong>Students are advised to practise on a daily basis, as it will help them achieve good marks on their board exams. The pdf of <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-24-the-circle\/\"><strong>RD Sharma Solutions Class 11 Maths Chapter 24<\/strong><\/a> Exercise 24.1 is available in the links below, which may be readily downloaded and saved for future reference.<\/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-69e750b99c55c\" 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\" 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href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-24-exercise-24-1\/#download-rd-sharma-class-11-solutions-chapter-24-exercise-241-free-pdf\" title=\"Download RD Sharma Class 11 Solutions Chapter 24 Exercise 24.1 Free PDF\">Download RD Sharma Class 11 Solutions Chapter 24 Exercise 24.1 Free PDF<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-24-exercise-24-1\/#access-answers-to-rd-sharma-solutions-class-11-maths-chapter-24-exercise-241-important-question-with-answers\" title=\"Access answers to RD Sharma Solutions Class 11 Maths Chapter 24 Exercise 24.1- Important Question with Answers\">Access answers to RD Sharma Solutions Class 11 Maths Chapter 24 Exercise 24.1- Important Question with Answers<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link 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href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-24-exercise-24-1\/#is-rd-sharma-solutions-for-class-12-maths-chapter-24-exercise-241-free\" title=\"Is RD Sharma Solutions for Class 12 Maths Chapter 24 Exercise 24.1 free?\">Is RD Sharma Solutions for Class 12 Maths Chapter 24 Exercise 24.1 free?<\/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-exercise-24-1\/#why-is-kopykitab%e2%80%99s-rd-sharma-solutions-class-11-maths-chapter-24-exercise-241-the-best-study-material\" title=\"Why is Kopykitab\u2019s RD Sharma Solutions Class 11 Maths Chapter 24 Exercise 24.1 the best study material?\">Why is Kopykitab\u2019s RD Sharma Solutions Class 11 Maths Chapter 24 Exercise 24.1 the best study material?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-class-11-solutions-chapter-24-exercise-241-free-pdf\"><\/span>Download RD Sharma Class 11 Solutions Chapter 24 Exercise 24.1 Free 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-Ex-24.1-1.pdf\">RD Sharma Solutions Class 11 Maths Chapter 24 Exercise 24.1<\/a><\/p>\n<div id=\"example1\" style=\"text-align: justify;\">\u00a0<\/div>\n<p style=\"text-align: justify;\"><style>\n.pdfobject-container { height: 800px;}<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-Ex-24.1-1.pdf\", \"#example1\");<\/script><\/p>\n<p>This exercise focuses on determining the equation of a circle when its center and radius are known, as well as determining the center and radius of a given circle. The <a href=\"https:\/\/www.kopykitab.com\/blog\/cbse-class-11-maths-rd-sharma-solutions\/\"><strong>RD Sharma Class 11 Solutions<\/strong><\/a> are offered here to assist students in achieving good results on their board exams. Our expert faculty team creates the solutions in a step-by-step format to ensure that students grasp the concepts.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"access-answers-to-rd-sharma-solutions-class-11-maths-chapter-24-exercise-241-important-question-with-answers\"><\/span>Access answers to RD Sharma Solutions Class 11 Maths Chapter 24 Exercise 24.1- Important Question with Answers<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<article id=\"post-53797\" class=\"post-53797 page type-page status-publish hentry\">\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 centre (-2, 3)<\/p>\n<p>By using the formula,<\/p>\n<p>The equation of the circle with centre (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>\u2234 The equation of the circle is x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0+ 4x \u2013 6y \u2013 3 = 0<\/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 centre (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>\u2234 The equation of the circle is x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0\u2013 2ax \u2013 2by = 0<\/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 centre (0, -1)<\/p>\n<p>By using the formula,<\/p>\n<p>The equation of the circle with centre (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>\u2234 The equation of the circle is x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0+ 2y = 0.<\/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 centre (a cos \u03b1, a sin \u03b1)<\/p>\n<p>By using the formula,<\/p>\n<p>The equation of the circle with centre (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>\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.<\/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 centre (a, a)<\/p>\n<p>By using the formula,<\/p>\n<p>The equation of the circle with centre (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>\u2234 The equation of the circle is x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0\u2013 2ax \u2013 2ay = 0.<\/p>\n<p><strong>2. Find the centre 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 centre 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>\u2234 The centre of the circle is (1, 0), and the radius is 2.<\/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 centre 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>\u2234 The centre of the circle is (-5, -1), and the radius is 3.<\/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 centre 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>\u2234 The centre of the circle is (2, -3), and the radius is 3\u221a2.<\/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 centre 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>\u2234 The centre of the circle is (\u00bd, -1), and the radius is \u221a17\/2.<\/p>\n<p><strong>3. Find the equation of the circle whose centre 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 the circle 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 centre (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>\u2234 The equation of the circle is x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0\u2013 2x \u2013 4y \u2013 20 = 0.<\/p>\n<p><strong>4. Find the equation of the circle passing through the point of intersection of the lines x + 3y = 0 and 2x \u2013 7y = 0 and whose centre is the point of intersection of the lines x + y + 1 = 0 and x \u2013 2y + 4 = 0.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let us find the points of intersection of the lines.<\/p>\n<p>On solving the lines x + 3y = 0 and 2x \u2013 7y = 0, we get the point of intersection to be (0, 0)<\/p>\n<p>On solving the lines x + y + 1 and x \u2013 2y + 4 = 0, we get the point of intersection to be (-2, 1)<\/p>\n<p>We have a circle with the centre (-2, 1) and passing through the point (0, 0).<\/p>\n<p>We know that the radius of the circle is the distance between the centre and any point on the radius. So, we find the radius of the circle.<\/p>\n<p>So, the equation 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 = 1<\/p>\n<p>(x + 2)<sup>2<\/sup>\u00a0+ (y \u2013 1)<sup>2<\/sup>\u00a0= r<sup>2<\/sup>\u00a0\u2026. (1)<\/p>\n<p>Equation (1) passes through (0, 0)<\/p>\n<p>So, (0 + 2)<sup>2<\/sup>\u00a0+ (0 \u2013 1)<sup>2<\/sup>\u00a0= r<sup>2<\/sup><\/p>\n<p>4 + 1 = a<sup>2<\/sup><\/p>\n<p>5 = r<sup>2<\/sup><\/p>\n<p>r = \u221a5<\/p>\n<p>We know that the equation of the circle with centre (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 (-2))<sup>2<\/sup>\u00a0+ (y \u2013 1)<sup>2<\/sup>\u00a0= (\u221a5)<sup>2<\/sup><\/p>\n<p>(x + 2)<sup>2<\/sup>\u00a0+ (y \u2013 1)<sup>2<\/sup>\u00a0= 5<\/p>\n<p>x<sup>2<\/sup>\u00a0+ 4x + 4 + y<sup>2<\/sup>\u00a0\u2013 2y + 1 = 5<\/p>\n<p>x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0+ 4x \u2013 2y = 0<\/p>\n<p>\u2234 The equation of the circle is x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0+ 4x \u2013 2y = 0.<\/p>\n<p><strong>5. Find the equation of the circle whose centre lies on the positive direction of y \u2013 axis at a distance 6 from the origin and whose radius is 4.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>It is given that the centre lies on the positive y-axis at a distance of 6 from the origin, we get the centre (0, 6).<\/p>\n<p>We have a circle with the centre at (0, 6) and radius 4.<\/p>\n<p>We know that the equation of the circle with centre (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>Where, p = 0, q = 6, r = 4<\/p>\n<p>Now by substituting the values in the equation, we get<\/p>\n<p>(x \u2013 0)<sup>2<\/sup>\u00a0+ (y \u2013 6)<sup>2<\/sup>\u00a0= 4<sup>2<\/sup><\/p>\n<p>x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0\u2013 12y + 36 = 16<\/p>\n<p>x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0\u2013 12y + 20 = 0.<\/p>\n<p>\u2234 The equation of the circle is x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0\u2013 12y + 20 = 0.<\/p>\n<p><strong>6. If the equations of two diameters of a circle are 2x + y = 6 and 3x + 2y = 4 and the radius is 10, find the equation of the circle.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>It is given that the circle has a radius 10 and has diameters 2x + y = 6 and 3x + 2y = 4.<\/p>\n<p>We know that the centre is the intersection point of the diameters.<\/p>\n<p>On solving the diameters, we get the centre to be (8, -10).<\/p>\n<p>We have a circle with the centre at (8, -10) and a radius of 10.<\/p>\n<p>By using the formula,<\/p>\n<p>We know that the equation of the circle with centre (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>Where, p = 8, q = -10, r = 10<\/p>\n<p>Now by substituting the values in the equation, we get<\/p>\n<p>(x \u2013 8)<sup>2<\/sup>\u00a0+ (y \u2013 (-10))<sup>2<\/sup>\u00a0= 10<sup>2<\/sup><\/p>\n<p>(x \u2013 8)<sup>2<\/sup>\u00a0+ (y + 10)<sup>2<\/sup>\u00a0= 100<\/p>\n<p>x<sup>2<\/sup>\u00a0\u2013 16x + 64 + y<sup>2<\/sup>\u00a0+ 20y + 100 = 100<\/p>\n<p>x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0\u2013 16x + 20y + 64 = 0.<\/p>\n<p>\u2234 The equation of the circle is x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0\u2013 16x + 20y + 64 = 0.<\/p>\n<p><strong>7.<\/strong>\u00a0<strong>Find the equation of the circle<br \/>(i) which touches both the axes at a distance of 6 units from the origin.<\/strong><\/p>\n<p><strong>(ii) Which touches x \u2013 axis at a distance of 5 from the origin and radius 6 units.<\/strong><\/p>\n<p><strong>(iii) Which touches both the axes and passes through the point (2, 1).<\/strong><\/p>\n<p><strong>(iv) Passing through the origin, radius 17 and ordinate of the centre is \u2013 15.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong>(i)\u00a0<\/strong>which touches both axes at a distance of 6 units from the origin.<\/p>\n<p>A circle\u00a0touches the axes at the points (\u00b16, 0) and (0, \u00b16).<\/p>\n<p>So, a circle has a centre (\u00b16, \u00b16) and passes through the point (0, 6).<\/p>\n<p>We know that the radius of the circle is the distance between the centre and any point on the radius. So, we find the radius of the circle.<\/p>\n<p>So, the equation is (x \u2013 p)<sup>2<\/sup>\u00a0+ (y \u2013 q)<sup>2<\/sup>\u00a0= r<sup>2<\/sup><\/p>\n<p>Where, p = 6, q = 6<\/p>\n<p>(x \u2013 6)<sup>2<\/sup>\u00a0+ (y \u2013 6)<sup>2<\/sup>\u00a0= r<sup>2<\/sup>\u00a0\u2026. (1)<\/p>\n<p>Equation (1) passes through (0, 6)<\/p>\n<p>So, (0 \u2013 6)<sup>2<\/sup>\u00a0+ (6 \u2013 6)<sup>2<\/sup>\u00a0= r<sup>2<\/sup><\/p>\n<p>36 + 0 = r<sup>2<\/sup><\/p>\n<p>r = \u221a36<\/p>\n<p>= 6<\/p>\n<p>We know that the equation of the circle with centre (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>Now by substituting the values in the equation, we get<\/p>\n<p>(x \u00b1 6)<sup>2<\/sup>\u00a0+ (y \u00b1 6)<sup>2<\/sup>\u00a0= (6)<sup>2<\/sup><\/p>\n<p>x<sup>2\u00a0<\/sup>\u00b1 12x + 36 + y<sup>2\u00a0<\/sup>\u00b1 12y + 36 = 36<\/p>\n<p>x<sup>2<\/sup>\u00a0+ y<sup>2\u00a0<\/sup>\u00b1 12x \u00b1 12y + 36 = 0<\/p>\n<p>\u2234 The equation of the circle is x<sup>2<\/sup>\u00a0+ y<sup>2\u00a0<\/sup>\u00b1 12x \u00b1 12y + 36 = 0.<\/p>\n<p><strong>(ii)\u00a0<\/strong>Which touches x-axis at a distance of 5 from the origin and radius 6 units.<\/p>\n<p>A circle touches the x-axis at the points (\u00b15, 0).<\/p>\n<p>Let us assume the centre of the circle is (\u00b15, a).<\/p>\n<p>We have a circle with the centre (5, a) and passing through the point (5, 0) and having a radius of 6.<\/p>\n<p>We know that the radius of the circle is the distance between the centre and any point on the radius. So, we find the radius of the circle.<\/p>\n<p>So, the equation is (x \u2013 p)<sup>2<\/sup>\u00a0+ (y \u2013 q)<sup>2<\/sup>\u00a0= r<sup>2<\/sup><\/p>\n<p>Where, p = 5, q = a<\/p>\n<p>(x \u2013 5)<sup>2<\/sup>\u00a0+ (y \u2013 a)<sup>2<\/sup>\u00a0= r<sup>2<\/sup>\u00a0\u2026. (1)<\/p>\n<p>Equation (1) passes through (5, 0)<\/p>\n<p>So, (5 \u2013 5)<sup>2<\/sup>\u00a0+ (0 \u2013 6)<sup>2<\/sup>\u00a0= r<sup>2<\/sup><\/p>\n<p>0 + 36 = r<sup>2<\/sup><\/p>\n<p>r = \u221a36<\/p>\n<p>= 6<\/p>\n<p>We have got the centre at (\u00b15, \u00b16) and has radius of 6 units.<\/p>\n<p>We know that the equation of the circle with centre (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>Now by substituting the values in the equation, we get<\/p>\n<p>(x\u00a0\u00b1\u00a05)<sup>2<\/sup>\u00a0+ (y \u00b1 6)<sup>2<\/sup>\u00a0= (6)<sup>2<\/sup><\/p>\n<p>x<sup>2<\/sup>\u00a0\u00b1 10x + 25 + y<sup>2<\/sup>\u00a0\u00b1 12y + 36 = 36<\/p>\n<p>x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0\u00b1 10x \u00b1 12y + 25 = 0.<\/p>\n<p>\u2234 The equation of the circle is x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0\u00b1 10x \u00b1 12y + 25 = 0.<\/p>\n<p><strong>(iii)\u00a0<\/strong>Which touches both the axes and passes through the point (2, 1).<\/p>\n<p>Let us assume the circle touches the x-axis at the point (a, 0) and the y-axis at the point (0, a).<\/p>\n<p>Then the centre of the circle is (a, a), and the radius is a.<\/p>\n<p>Its equation will be (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, we get<\/p>\n<p>(x \u2013 a)<sup>2<\/sup>\u00a0+ (y \u2013 a)<sup>2<\/sup>\u00a0= a<sup>2<\/sup>\u00a0\u2026 (1)<\/p>\n<p>So now, equation (1) passes through P (2, 1)<\/p>\n<p>By substituting the values, we get<\/p>\n<p>(2 \u2013 a)<sup>2<\/sup>\u00a0+ (1 \u2013 a)<sup>2<\/sup>\u00a0= a<sup>2<\/sup><\/p>\n<p>4 \u2013 4a + a<sup>2<\/sup>\u00a0+ 1 \u2013 2a + a<sup>2<\/sup>\u00a0= a<sup>2<\/sup><\/p>\n<p>5 \u2013 6a + a<sup>2<\/sup>\u00a0= 0<\/p>\n<p>(a \u2013 5) (a \u2013 1) = 0<\/p>\n<p>So, a = 5 or 1<\/p>\n<p>Case (i)<\/p>\n<p>We have got the centre at (5, 5) and has a radius of 5 units.<\/p>\n<p>We know that the equation of the circle with centre (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>Now by substituting the values in the equation, we get<\/p>\n<p>(x \u2013 5)<sup>2<\/sup>\u00a0+ (y \u2013 5)<sup>2<\/sup>\u00a0= 5<sup>2<\/sup><\/p>\n<p>x<sup>2<\/sup>\u00a0\u2013 10x + 25 + y<sup>2<\/sup>\u00a0\u2013 10y + 25 = 25<\/p>\n<p>x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0\u2013 10x \u2013 10y + 25 = 0.<\/p>\n<p>\u2234 The equation of the circle is x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0\u2013 10x \u2013 10y + 25 = 0.<\/p>\n<p>Case (ii)<\/p>\n<p>We have got the centre at (1, 1) and having a radius of 1 unit.<\/p>\n<p>We know that the equation of the circle with centre (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>Now by substituting the values in the equation, we get<\/p>\n<p>(x \u2013 1)<sup>2<\/sup>\u00a0+ (y \u2013 1)<sup>2<\/sup>\u00a0= 1<sup>2<\/sup><\/p>\n<p>x<sup>2<\/sup>\u00a0\u2013 2x + 1 + y<sup>2<\/sup>\u00a0\u2013 2y + 1 = 1<\/p>\n<p>x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0\u2013 2x \u2013 2y + 1 = 0<\/p>\n<p>\u2234\u00a0The equation of the circle is x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0\u2013 2x \u2013 2y + 1 = 0.<\/p>\n<p><strong>(iv)\u00a0<\/strong>Passing through the origin, radius 17 and ordinate of the centre is \u2013 15.<\/p>\n<p>Let us assume the abscissa as \u2018a\u2019<\/p>\n<p>We have a circle with a centre (a, \u2013 15) and passing through the point (0, 0) and having a radius of 17.<\/p>\n<p>We know that the radius of the circle is the distance between the centre and any point on the radius. So, we find the radius of the circle.<\/p>\n<p>By using the distance formula,<\/p>\n<p><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-6.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 24 \u2013 The Circle - image 6\" \/><\/p>\n<p>17<sup>2<\/sup>\u00a0= a<sup>2<\/sup>\u00a0+ (-15)<sup>2<\/sup><\/p>\n<p>289 = a<sup>2<\/sup>\u00a0+ 225<\/p>\n<p>a<sup>2<\/sup>\u00a0= 64<\/p>\n<p>|a| = \u221a64<\/p>\n<p>|a| = 8<\/p>\n<p>a = \u00b18 \u2026. (1)<\/p>\n<p>We have got the centre at (\u00b18, \u2013 15) and has radius of 17 units.<\/p>\n<p>We know that the equation of the circle with centre (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>Now by substituting the values in the equation, we get<\/p>\n<p>(x \u00b1 8)<sup>2<\/sup>\u00a0+ (y \u2013 15)<sup>2<\/sup>\u00a0= 17<sup>2<\/sup><\/p>\n<p>x<sup>2\u00a0<\/sup>\u00b1 16x + 64 + y<sup>2<\/sup>\u00a0\u2013 30y + 225 = 289<\/p>\n<p>x<sup>2<\/sup>\u00a0+ y<sup>2\u00a0<\/sup>\u00b1 16x \u2013 30y = 0.<\/p>\n<p>\u2234 The equation of the circle is x<sup>2<\/sup>\u00a0+ y<sup>2\u00a0<\/sup>\u00b1 16x \u2013 30y = 0.<\/p>\n<p><strong>8. Find the equation of the circle which has its centre at the point (3, 4) and touches the straight line 5x + 12y \u2013 1 = 0.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>It is given that we need to find the equation of the circle with centre (3, 4) and touches the straight line 5x + 12y \u2013 1 = 0.<\/p>\n<p><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-7.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 24 \u2013 The Circle - image 7\" \/><\/p>\n<p><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-8.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 24 \u2013 The Circle - image 8\" \/><\/p>\n<p>We have a circle with centre (3, 4) and a radius of 62\/13.<\/p>\n<p>We know that the equation of the circle with centre (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>Now by substituting the values in the equation, we get<\/p>\n<p>(x \u2013 3)<sup>2<\/sup>\u00a0+ (y \u2013 4)<sup>2<\/sup>\u00a0= (62\/13)<sup>2<\/sup><\/p>\n<p>x<sup>2<\/sup>\u00a0\u2013 6x + 9 + y<sup>2<\/sup>\u00a0\u2013 8y + 16 = 3844\/169<\/p>\n<p>169x<sup>2<\/sup>\u00a0+ 169y<sup>2<\/sup>\u00a0\u2013 1014x \u2013 1352y + 4225 = 3844<\/p>\n<p>169x<sup>2<\/sup>\u00a0+ 169y<sup>2<\/sup>\u00a0\u2013 1014x \u2013 1352y + 381 = 0<\/p>\n<p>\u2234 The equation of the circle is 169x<sup>2<\/sup>\u00a0+ 169y<sup>2<\/sup>\u00a0\u2013 1014x \u2013 1352y + 381 = 0.<\/p>\n<p><strong>9. Find the equation of the circle which touches the axes and whose centre lies on x \u2013 2y = 3.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let us assume the circle touches the axes at (a, 0) and (0, a), and we get the radius to be |a|.<\/p>\n<p>We get the centre of the circle as (a, a). This point lies on the line x \u2013 2y = 3<\/p>\n<p>a \u2013 2(a) = 3<\/p>\n<p>-a = 3<\/p>\n<p>a = \u2013 3<\/p>\n<p>Centre = (a, a) = (-3, -3) and radius of the circle(r) = |-3| = 3<\/p>\n<p>We have a circle with the centre at (-3, -3) and a radius of 3.<\/p>\n<p>We know that the equation of the circle with centre (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>Now by substituting the values in the equation, we get<\/p>\n<p>(x \u2013 (-3))<sup>2<\/sup>\u00a0+ (y \u2013 (-3))<sup>2<\/sup>\u00a0= 3<sup>2<\/sup><\/p>\n<p>(x + 3)<sup>2<\/sup>\u00a0+ (y + 3)<sup>2<\/sup>\u00a0= 9<\/p>\n<p>x<sup>2<\/sup>\u00a0+ 6x + 9 + y<sup>2<\/sup>\u00a0+ 6y + 9 = 9<\/p>\n<p>x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0+ 6x + 6y + 9 = 0<\/p>\n<p>\u2234 The equation of the circle is x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0+ 6x + 6y + 9 = 0.<\/p>\n<p><strong>10. A circle whose centre is the point of intersection of the lines 2x \u2013 3y + 4 = 0 and 3x + 4y \u2013 5 = 0 passes through the origin. Find its equation.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>It is given that the circle has the centre at the intersection point of the lines 2x \u2013 3y + 4 = 0 and 3x + 4y \u2013 5 = 0 and passes through the origin<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-9.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 24 \u2013 The Circle - image 9\" \/><\/p>\n<p>We know that the equation of the circle with centre (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>Now by substituting the values in the equation, we get<\/p>\n<p><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-10.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 24 \u2013 The Circle - image 10\" \/><\/p>\n<\/article>\n<p>We have provided complete details of RD Sharma Solutions Class 11 Maths Chapter 24 Exercise 24.1. If you have any queries related to <a href=\"https:\/\/www.cbse.gov.in\/\" target=\"_blank\" rel=\"noopener\"><strong>CBSE<\/strong><\/a>\u00a0Class 11, feel free to ask us in the comment section below.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-class-11-solutions-chapter-24-exercise-241\"><\/span>FAQs on RD Sharma Class 11 Solutions Chapter 24 Exercise 24.1<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-1630585388047\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"where-can-i-download-rd-sharma-class-11-solutions-chapter-24-exercise-241-free-pdf\"><\/span>Where can I download RD Sharma Class 11 Solutions Chapter 24 Exercise 24.1 free PDF?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can download RD Sharma Solutions for Class 11 Maths Chapter 24 Exercise 24.1 free PDF from the above article.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630585452458\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"is-rd-sharma-solutions-for-class-12-maths-chapter-24-exercise-241-free\"><\/span>Is RD Sharma Solutions for Class 12 Maths Chapter 24 Exercise 24.1 free?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Yes, You can get RD Sharma Solutions Class 12 Maths Chapter 24 Exercise 24.1 for free.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630585493064\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"why-is-kopykitab%e2%80%99s-rd-sharma-solutions-class-11-maths-chapter-24-exercise-241-the-best-study-material\"><\/span>Why is Kopykitab\u2019s RD Sharma Solutions Class 11 Maths Chapter 24 Exercise 24.1 the best study material?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>The RD Sharma Solutions Class 11 Maths Chapter 24 Exercise 24.1 available on Kopykitab\u2019s website has been created by highly qualified experts to assist students in achieving high scores on the board exam. The solutions are well-organized and logical, giving pupils a clear picture of the most important questions.\u00a0<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Solutions Class 11 Maths Chapter 24 Exercise 24.1:\u00a0Students are advised to practise on a daily basis, as it will help them achieve good marks on their board exams. The pdf of RD Sharma Solutions Class 11 Maths Chapter 24 Exercise 24.1 is available in the links below, which may be readily downloaded and &#8230; <a title=\"RD Sharma Class 11 Solutions Chapter 24 Exercise 24.1 (Updated for 2023)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-24-exercise-24-1\/\" aria-label=\"More on RD Sharma Class 11 Solutions Chapter 24 Exercise 24.1 (Updated for 2023)\">Read more<\/a><\/p>\n","protected":false},"author":238,"featured_media":123795,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73411,2985,73410],"tags":[3428,73334,73564,4388],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/68067"}],"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\/238"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=68067"}],"version-history":[{"count":4,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/68067\/revisions"}],"predecessor-version":[{"id":377423,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/68067\/revisions\/377423"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/123795"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=68067"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=68067"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=68067"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}