{"id":106669,"date":"2023-05-23T12:07:00","date_gmt":"2023-05-23T06:37:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=106669"},"modified":"2025-07-12T11:55:04","modified_gmt":"2025-07-12T06:25:04","slug":"ncert-solutions-for-class-11-maths-chapter-11-exercise-11-1","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/ncert-solutions-for-class-11-maths-chapter-11-exercise-11-1\/","title":{"rendered":"NCERT Solutions Class 11 Maths Chapter 11 Exercise 11.1"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-146042\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/10\/11.1-1.jpg\" alt=\"NCERT Solutions for Class 11 Maths Chapter 11 Exercise 11.1\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/10\/11.1-1.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/10\/11.1-1-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>NCERT Solutions for Class 11 Maths Chapter 11 Exercise 11.1<\/strong>: All exercise 11.1 questions are answered in the NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections. The goal of practicing NCERT Solutions is to improve your final test result. To obtain the solutions to all of the exercises in all of the chapters, download NCERT Class 11 maths solutions.<\/p>\n<p>A conic section is a chapter that covers various cone sections. In its first exercise, it covers a wide range of themes. The following subjects are covered in Exercise 11.1 of <a href=\"https:\/\/www.kopykitab.com\/blog\/ncert-solutions-for-class-11-maths-chapter-11-conic-sections\/\"><strong>NCERT Solutions for Class 11 Maths Chapter 11- Conic Sections<\/strong><\/a>:<\/p>\n<ul>\n<li>Introduction<\/li>\n<li>Sections of a Cone\n<ul>\n<li>Circle, ellipse, parabola, and hyperbola<\/li>\n<li>Degenerated conic sections<\/li>\n<\/ul>\n<\/li>\n<li>Circle<\/li>\n<\/ul>\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-69ed0ff01ac42\" aria-label=\"Table of Content\"><span style=\"display: flex;align-items: center;width: 35px;height: 30px;justify-content: 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ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/www.kopykitab.com\/blog\/ncert-solutions-for-class-11-maths-chapter-11-exercise-11-1\/#download-ncert-solutions-for-class-11-maths-chapter-11-exercise-111-free-pdf\" title=\"Download NCERT Solutions for Class 11 Maths Chapter 11 Exercise 11.1 Free PDF\">Download NCERT Solutions for Class 11 Maths Chapter 11 Exercise 11.1 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\/ncert-solutions-for-class-11-maths-chapter-11-exercise-11-1\/#access-the-ncert-solutions-for-class-11-maths-chapter-11-%e2%80%93-exercise-111\" title=\"Access the NCERT Solutions for Class 11 Maths Chapter 11 \u2013 Exercise 11.1\">Access the NCERT Solutions for Class 11 Maths Chapter 11 \u2013 Exercise 11.1<\/a><\/li><\/ul><\/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\/ncert-solutions-for-class-11-maths-chapter-11-exercise-11-1\/#faqs-on-ncert-solutions-for-class-11-maths-chapter-11-exercise-111\" title=\"FAQs on NCERT Solutions for Class 11 Maths Chapter 11 Exercise 11.1\">FAQs on NCERT Solutions for Class 11 Maths Chapter 11 Exercise 11.1<\/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\/ncert-solutions-for-class-11-maths-chapter-11-exercise-11-1\/#what-is-chapter-11-of-ncert-maths-of-class-11\" title=\"What is Chapter 11 of NCERT Maths of Class 11?\">What is Chapter 11 of NCERT Maths of Class 11?<\/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\/ncert-solutions-for-class-11-maths-chapter-11-exercise-11-1\/#where-can-i-get-class-11-maths-ncert-solutions-for-chapter-11-exercise-111\" title=\"Where can I get Class 11 Maths NCERT Solutions for Chapter 11 Exercise 11.1?\">Where can I get Class 11 Maths NCERT Solutions for Chapter 11 Exercise 11.1?<\/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\/ncert-solutions-for-class-11-maths-chapter-11-exercise-11-1\/#how-many-exercises-are-there-in-chapter-11-of-maths-ncert-of-class-11\" title=\"How many exercises are there in Chapter 11 of Maths NCERT of Class 11?\">How many exercises are there in Chapter 11 of Maths NCERT of Class 11?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-7\" href=\"https:\/\/www.kopykitab.com\/blog\/ncert-solutions-for-class-11-maths-chapter-11-exercise-11-1\/#what-are-some-of-the-benefits-of-using-ncert-math-solutions\" title=\"What are some of the benefits of using NCERT Math Solutions?\">What are some of the benefits of using NCERT Math Solutions?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-8\" href=\"https:\/\/www.kopykitab.com\/blog\/ncert-solutions-for-class-11-maths-chapter-11-exercise-11-1\/#what-is-exercise-111-of-the-ncert-maths-of-class-11-chapter-11-based-on\" title=\"What is exercise 11.1 of the NCERT Maths of Class 11 Chapter 11 based on?\">What is exercise 11.1 of the NCERT Maths of Class 11 Chapter 11 based on?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-9\" href=\"https:\/\/www.kopykitab.com\/blog\/ncert-solutions-for-class-11-maths-chapter-11-exercise-11-1\/#what-are-ncert-maths-solutions-for-class-11\" title=\"What are NCERT Maths Solutions for Class 11?\">What are NCERT Maths Solutions for Class 11?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-ncert-solutions-for-class-11-maths-chapter-11-exercise-111-free-pdf\"><\/span><strong>Download NCERT Solutions for Class 11 Maths Chapter 11 Exercise 11.1 Free PDF<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<div id=\"example1\" style=\"text-align: justify;\">&nbsp;<\/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\/07\/ncert-jan2021-solution-class-11-chapter-11-conic-ex-1.pdf\", \"#example1\");<\/script><\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/07\/ncert-jan2021-solution-class-11-chapter-11-conic-ex-1.pdf\">NCERT Solutions for Class 11 Maths Chapter 11 Exercise 11.1<\/a><\/p>\n<h3><span class=\"ez-toc-section\" id=\"access-the-ncert-solutions-for-class-11-maths-chapter-11-%e2%80%93-exercise-111\"><\/span>Access the <strong>NCERT <\/strong>Solutions for Class 11 Maths Chapter 11 \u2013 Exercise 11.1<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>In each of the following Exercises 1 to 5, find the equation of the circle with<br>1. center (0, 2) and radius 2.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>center (0, 2) and radius 2.<\/p>\n<p>Let us consider the equation of a circle with center (h, k), and<\/p>\n<p>Radius r is given as (x \u2013 h)<sup>2&nbsp;<\/sup>+ (y \u2013 k)<sup>2&nbsp;<\/sup>= r<sup>2<\/sup><\/p>\n<p>So, center (h, k) = (0, 2) and radius (r) = 2<\/p>\n<p>The equation of the circle is<\/p>\n<p>(x \u2013 0)<sup>2&nbsp;<\/sup>+ (y \u2013 2)<sup>2&nbsp;<\/sup>= 2<sup>2<\/sup><\/p>\n<p>x<sup>2&nbsp;<\/sup>+ y<sup>2&nbsp;<\/sup>+ 4 \u2013 4y = 4<\/p>\n<p>x<sup>2&nbsp;<\/sup>+ y<sup>2<\/sup>&nbsp;\u2013 4y = 0<\/p>\n<p>\u2234 The equation of the circle is x<sup>2&nbsp;<\/sup>+ y<sup>2<\/sup>&nbsp;\u2013 4y = 0<\/p>\n<p><strong>2. center (\u20132, 3) and radius 4.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>center (-2, 3) and radius 4<\/p>\n<p>Let us consider the equation of a circle with center (h, k), and<\/p>\n<p>Radius r is given as (x \u2013 h)<sup>2&nbsp;<\/sup>+ (y \u2013 k)<sup>2&nbsp;<\/sup>= r<sup>2<\/sup><\/p>\n<p>So, center (h, k) = (-2, 3) and radius (r) = 4<\/p>\n<p>The equation of the circle is<\/p>\n<p>(x + 2)<sup>2<\/sup>&nbsp;+ (y \u2013 3)<sup>2<\/sup>&nbsp;= (4)<sup>2<\/sup><\/p>\n<p>x<sup>2<\/sup>&nbsp;+ 4x + 4 + y<sup>2<\/sup>&nbsp;\u2013 6y + 9 = 16<\/p>\n<p>x<sup>2<\/sup>&nbsp;+ y<sup>2<\/sup>&nbsp;+ 4x \u2013 6y \u2013 3 = 0<\/p>\n<p>\u2234 The equation of the circle is x<sup>2<\/sup>&nbsp;+ y<sup>2<\/sup>&nbsp;+ 4x \u2013 6y \u2013 3 = 0<\/p>\n<p><strong>3. center (1\/2, 1\/4) and radius (1\/12).<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>center (1\/2, 1\/4) and radius 1\/12<\/p>\n<p>Let us consider the equation of a circle with center (h, k), and<\/p>\n<p>Radius r is given as (x \u2013 h)<sup>2&nbsp;<\/sup>+ (y \u2013 k)<sup>2&nbsp;<\/sup>= r<sup>2<\/sup><\/p>\n<p>So, center (h, k) = (1\/2, 1\/4) and radius (r) = 1\/12<\/p>\n<p>The equation of the circle is<\/p>\n<p>(x \u2013 1\/2)<sup>2<\/sup>&nbsp;+ (y \u2013 1\/4)<sup>2<\/sup>&nbsp;= (1\/12)<sup>2<\/sup><\/p>\n<p>x<sup>2<\/sup>&nbsp;\u2013 x + \u00bc + y<sup>2<\/sup>&nbsp;\u2013 y\/2 + 1\/16 = 1\/144<\/p>\n<p>x<sup>2<\/sup>&nbsp;\u2013 x + \u00bc + y<sup>2<\/sup>&nbsp;\u2013 y\/2 + 1\/16 = 1\/144<\/p>\n<p>144x<sup>2<\/sup>&nbsp;\u2013 144x + 36 + 144y<sup>2<\/sup>&nbsp;\u2013 72y + 9 \u2013 1 = 0<\/p>\n<p>144x<sup>2<\/sup>&nbsp;\u2013 144x + 144y<sup>2<\/sup>&nbsp;\u2013 72y + 44 = 0<\/p>\n<p>36x<sup>2<\/sup>&nbsp;+ 36x + 36y<sup>2<\/sup>&nbsp;\u2013 18y + 11 = 0<\/p>\n<p>36x<sup>2<\/sup>&nbsp;+ 36y<sup>2<\/sup>&nbsp;\u2013 36x \u2013 18y + 11= 0<\/p>\n<p>\u2234 The equation of the circle is 36x<sup>2<\/sup>&nbsp;+ 36y<sup>2<\/sup>&nbsp;\u2013 36x \u2013 18y + 11= 0<\/p>\n<p><strong>4. center (1, 1) and radius \u221a2<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>center (1, 1) and radius \u221a2<\/p>\n<p>Let us consider the equation of a circle with center (h, k), and<\/p>\n<p>Radius r is given as (x \u2013 h)<sup>2&nbsp;<\/sup>+ (y \u2013 k)<sup>2&nbsp;<\/sup>= r<sup>2<\/sup><\/p>\n<p>So, center (h, k) = (1, 1) and radius (r) = \u221a2<\/p>\n<p>The equation of the circle is<\/p>\n<p>(x-1)<sup>2<\/sup>&nbsp;+ (y-1)<sup>2&nbsp;<\/sup>= (\u221a2)<sup>2<\/sup><\/p>\n<p>x<sup>2<\/sup>&nbsp;\u2013 2x + 1 + y<sup>2<\/sup>&nbsp;-2y + 1 = 2<\/p>\n<p>x<sup>2<\/sup>&nbsp;+ y<sup>2<\/sup>&nbsp;\u2013 2x -2y = 0<\/p>\n<p>\u2234 The equation of the circle is x<sup>2<\/sup>&nbsp;+ y<sup>2<\/sup>&nbsp;\u2013 2x -2y = 0<\/p>\n<p><strong>5. center (\u2013a, \u2013b) and radius \u221a(a<sup>2<\/sup>&nbsp;\u2013 b<sup>2<\/sup>)<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>center (-a, -b) and radius \u221a(a<sup>2<\/sup>&nbsp;\u2013 b<sup>2<\/sup>)<\/p>\n<p>Let us consider the equation of a circle with center (h, k), and<\/p>\n<p>Radius r is given as (x \u2013 h)<sup>2&nbsp;<\/sup>+ (y \u2013 k)<sup>2&nbsp;<\/sup>= r<sup>2<\/sup><\/p>\n<p>So, center (h, k) = (-a, -b) and radius (r) = \u221a(a<sup>2<\/sup>&nbsp;\u2013 b<sup>2<\/sup>)<\/p>\n<p>The equation of the circle is<\/p>\n<p>(x + a)<sup>2<\/sup>&nbsp;+ (y + b)<sup>2<\/sup>&nbsp;= (\u221a(a<sup>2<\/sup>&nbsp;\u2013 b<sup>2<\/sup>)<sup>2<\/sup>)<\/p>\n<p>x<sup>2<\/sup>&nbsp;+ 2ax + a<sup>2<\/sup>&nbsp;+ y<sup>2<\/sup>&nbsp;+ 2by + b<sup>2<\/sup>&nbsp;= a<sup>2<\/sup>&nbsp;\u2013 b<sup>2<\/sup><\/p>\n<p>x<sup>2<\/sup>&nbsp;+ y<sup>2<\/sup>&nbsp;+2ax + 2by + 2b<sup>2<\/sup>&nbsp;= 0<\/p>\n<p>\u2234 The equation of the circle is x<sup>2<\/sup>&nbsp;+ y<sup>2<\/sup>&nbsp;+2ax + 2by + 2b<sup>2<\/sup>&nbsp;= 0<\/p>\n<p><strong>In each of the following Exercises 6 to 9, find the center and radius of the circles.<\/strong><\/p>\n<p>6. (x + 5)<sup>2<\/sup>&nbsp;+ (y \u2013 3)<sup>2<\/sup>&nbsp;= 36<\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>The equation of the given circle is (x + 5)<sup>2<\/sup>&nbsp;+ (y \u2013 3)<sup>2<\/sup>&nbsp;= 36<\/p>\n<p>(x \u2013 (-5))<sup>2<\/sup>&nbsp;+ (y \u2013 3)<sup>2<\/sup>&nbsp;= 6<sup>2<\/sup>&nbsp;[which is of the form (x \u2013 h)<sup>2<\/sup>&nbsp;+ (y \u2013 k )<sup>2<\/sup>&nbsp;= r<sup>2<\/sup>]<\/p>\n<p>Where, h = -5, k = 3 and r = 6<\/p>\n<p>\u2234 The center of the given circle is (-5, 3), and its radius is 6.<\/p>\n<p><strong>7. x<sup>2<\/sup>&nbsp;+ y<sup>2<\/sup>&nbsp;\u2013 4x \u2013 8y \u2013 45 = 0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>The equation of the given circle is x<sup>2<\/sup>&nbsp;+ y<sup>2<\/sup>&nbsp;\u2013 4x \u2013 8y \u2013 45 = 0<\/p>\n<p>x<sup>2<\/sup>&nbsp;+ y<sup>2<\/sup>&nbsp;\u2013 4x \u2013 8y \u2013 45 = 0<\/p>\n<p>(x<sup>2<\/sup>&nbsp;\u2013 4x) + (y<sup>2<\/sup>&nbsp;-8y) = 45<\/p>\n<p>(x<sup>2<\/sup>&nbsp;\u2013 2(x) (2) + 2<sup>2<\/sup>) + (y<sup>2<\/sup>&nbsp;\u2013 2(y) (4) + 4<sup>2<\/sup>) \u2013 4 \u2013 16 = 45<\/p>\n<p>(x \u2013 2)<sup>2<\/sup>&nbsp;+ (y \u2013 4)<sup>2<\/sup>&nbsp;= 65<\/p>\n<p>(x \u2013 2)<sup>2<\/sup>&nbsp;+ (y \u2013 4)<sup>2<\/sup>&nbsp;= (\u221a65)<sup>2<\/sup>&nbsp;[which is in the form (x-h)<sup>2<\/sup>&nbsp;+(y-k)<sup>2<\/sup>&nbsp;= r<sup>2<\/sup>]<\/p>\n<p>Where h = 2, K = 4 and r = \u221a65<\/p>\n<p>\u2234 The center of the given circle is (2, 4), and its radius is \u221a65.<\/p>\n<p><strong>8. x<sup>2<\/sup>&nbsp;+ y<sup>2<\/sup>&nbsp;\u2013 8x + 10y \u2013 12 = 0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>The equation of the given circle is x<sup>2<\/sup>&nbsp;+ y<sup>2<\/sup>&nbsp;-8x + 10y -12 = 0<\/p>\n<p>x<sup>2<\/sup>&nbsp;+ y<sup>2<\/sup>&nbsp;\u2013 8x + 10y \u2013 12 = 0<\/p>\n<p>(x<sup>2<\/sup>&nbsp;\u2013 8x) + (y<sup>2&nbsp;<\/sup>+ 10y) = 12<\/p>\n<p>(x<sup>2<\/sup>&nbsp;\u2013 2(x) (4) + 4<sup>2<\/sup>) + (y<sup>2<\/sup>&nbsp;\u2013 2(y) (5) + 5<sup>2<\/sup>) \u2013 16 \u2013 25 = 12<\/p>\n<p>(x \u2013 4)<sup>2<\/sup>&nbsp;+ (y + 5)<sup>2<\/sup>&nbsp;= 53<\/p>\n<p>(x \u2013 4)<sup>2<\/sup>&nbsp;+ (y \u2013 (-5))<sup>2<\/sup>&nbsp;= (\u221a53)<sup>2<\/sup>&nbsp;[which is in the form (x-h)<sup>2<\/sup>&nbsp;+(y-k)<sup>2<\/sup>&nbsp;= r<sup>2<\/sup>]<\/p>\n<p>Where h = 4, K= -5 and r = \u221a53<\/p>\n<p>\u2234 The center of the given circle is (4, -5), and its radius is \u221a53.<\/p>\n<p><strong>9. 2x<sup>2<\/sup>&nbsp;+ 2y<sup>2<\/sup>&nbsp;\u2013 x = 0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>The equation of the given circle is 2x<sup>2<\/sup>&nbsp;+ 2y<sup>2<\/sup>&nbsp;\u2013x = 0<\/p>\n<p>2x<sup>2<\/sup>&nbsp;+ 2y<sup>2<\/sup>&nbsp;\u2013x = 0<\/p>\n<p>(2x<sup>2<\/sup>&nbsp;+ x) + 2y<sup>2<\/sup>&nbsp;= 0<\/p>\n<p>(x<sup>2<\/sup>&nbsp;\u2013 2 (x) (1\/4) + (1\/4)<sup>2<\/sup>) + y<sup>2<\/sup>&nbsp;\u2013 (1\/4)<sup>2<\/sup>&nbsp;= 0<\/p>\n<p>(x \u2013 1\/4)<sup>2<\/sup>&nbsp;+ (y \u2013 0)<sup>2<\/sup>&nbsp;= (1\/4)<sup>2<\/sup>&nbsp;[which is in the form (x-h)<sup>2<\/sup>&nbsp;+(y-k)<sup>2<\/sup>&nbsp;= r<sup>2<\/sup>]<\/p>\n<p>Where, h = \u00bc, K = 0, and r = \u00bc<\/p>\n<p>\u2234 The center of the given circle is (1\/4, 0), and its radius is 1\/4.<\/p>\n<p><strong>10. Find the equation of the circle passing through the points (4,1) and (6,5) and whose center is on line 4x + y = 16.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let us consider the equation of the required circle be (x \u2013 h)<sup>2<\/sup>+ (y \u2013 k)<sup>2<\/sup>&nbsp;= r<sup>2<\/sup><\/p>\n<p>We know that the circle passes through points (4,1) and (6,5).<\/p>\n<p>So,<\/p>\n<p>(4 \u2013 h)<sup>2&nbsp;<\/sup>+ (1 \u2013 k)<sup>2<\/sup>&nbsp;= r<sup>2<\/sup>&nbsp;\u2026\u2026\u2026\u2026\u2026..(1)<\/p>\n<p>(6\u2013 h)<sup>2<\/sup>+ (5 \u2013 k)<sup>2<\/sup>&nbsp;= r<sup>2<\/sup>&nbsp;\u2026\u2026\u2026\u2026\u2026\u2026(2)<\/p>\n<p>The center (h, k) of the circle lies on line 4x + y = 16.<\/p>\n<p>4h + k =16\u2026\u2026\u2026\u2026\u2026\u2026\u2026 (3)<\/p>\n<p>From the equation (1) and (2), we obtain<\/p>\n<p>(4 \u2013 h)<sup>2<\/sup>+ (1 \u2013 k)<sup>2<\/sup>&nbsp;=(6 \u2013 h)<sup>2<\/sup>&nbsp;+ (5 \u2013 k)<sup>2<\/sup><\/p>\n<p>16 \u2013 8h + h<sup>2<\/sup>&nbsp;+1 -2k +k<sup>2<\/sup>&nbsp;= 36 -12h +h<sup>2<\/sup>+15 \u2013 10k + k<sup>2<\/sup><\/p>\n<p>16 \u2013 8h +1 -2k + 12h -25 -10k<\/p>\n<p>4h +8k = 44<\/p>\n<p>h + 2k =11\u2026\u2026\u2026\u2026\u2026. (4)<\/p>\n<p>On solving equations (3) and (4), we obtain h=3 and k= 4.<\/p>\n<p>On substituting the values of h and k in equation (1), we obtain<\/p>\n<p>(4 \u2013 3)<sup>2<\/sup>+ (1 \u2013 4)<sup>2<\/sup>&nbsp;= r<sup>2<\/sup><\/p>\n<p>(1)<sup>2<\/sup>&nbsp;+ (-3)<sup>2<\/sup>&nbsp;= r<sup>2<\/sup><\/p>\n<p>1+9 = r<sup>2<\/sup><\/p>\n<p>r =&nbsp;\u221a10<\/p>\n<p>So now, (x \u2013 3)<sup>2&nbsp;<\/sup>+ (y \u2013 4)<sup>2<\/sup>&nbsp;= (\u221a10)<sup>2<\/sup><\/p>\n<p>x<sup>2<\/sup>&nbsp;\u2013 6x + 9 + y<sup>2<\/sup>&nbsp;\u2013 8y + 16 =10<\/p>\n<p>x<sup>2<\/sup>&nbsp;+ y<sup>2<\/sup>&nbsp;\u2013 6x \u2013 8y + 15 = 0<\/p>\n<p>\u2234 The equation of the required circle is x<sup>2<\/sup>&nbsp;+ y<sup>2<\/sup>&nbsp;\u2013 6x \u2013 8y + 15 = 0<\/p>\n<p><strong>11. Find the equation of the circle passing through the points (2, 3) and (\u20131, 1) and whose center is on the line x \u2013 3y \u2013 11 = 0.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let us consider the equation of the required circle be (x \u2013 h)<sup>2&nbsp;<\/sup>+ (y \u2013 k)<sup>2<\/sup>&nbsp;= r<sup>2<\/sup><\/p>\n<p>We know that the circle passes through points (2,3) and (-1,1).<\/p>\n<p>(2 \u2013 h)<sup>2<\/sup>+ (3 \u2013 k)<sup>2<\/sup>&nbsp;=r<sup>2<\/sup>&nbsp;\u2026\u2026\u2026\u2026\u2026..(1)<\/p>\n<p>(-1 \u2013 h)<sup>2<\/sup>+ (1\u2013 k)<sup>2<\/sup>&nbsp;=r<sup>2<\/sup>&nbsp;\u2026\u2026\u2026\u2026\u2026\u2026(2)<\/p>\n<p>The center (h, k) of the circle lies on line x \u2013 3y \u2013 11= 0.<\/p>\n<p>h \u2013 3k =11\u2026\u2026\u2026\u2026\u2026\u2026\u2026 (3)<\/p>\n<p>From the equation (1) and (2), we obtain<\/p>\n<p>(2 \u2013 h)<sup>2<\/sup>+ (3 \u2013 k)<sup>2<\/sup>&nbsp;=(-1 \u2013 h)<sup>2<\/sup>&nbsp;+ (1 \u2013 k)<sup>2<\/sup><\/p>\n<p>4 \u2013 4h + h<sup>2<\/sup>&nbsp;+9 -6k +k<sup>2<\/sup>&nbsp;= 1 + 2h +h<sup>2<\/sup>+1 \u2013 2k + k<sup>2<\/sup><\/p>\n<p>4 \u2013 4h +9 -6k = 1 + 2h + 1 -2k<\/p>\n<p>6h + 4k =11\u2026\u2026\u2026\u2026\u2026. (4)<\/p>\n<p>Now, let us multiply equation (3) by 6 and subtract it from equation (4) to get<\/p>\n<p>6h+ 4k \u2013 6(h-3k) = 11 \u2013 66<\/p>\n<p>6h + 4k \u2013 6h + 18k = 11 \u2013 66<\/p>\n<p>22 k = \u2013 55<\/p>\n<p>K = -5\/2<\/p>\n<p>Substitute this value of K in equation (4) to get<\/p>\n<p>6h + 4(-5\/2) = 11<\/p>\n<p>6h \u2013 10 = 11<\/p>\n<p>6h = 21<\/p>\n<p>h = 21\/6<\/p>\n<p>h = 7\/2<\/p>\n<p>We obtain h =&nbsp;7\/2and k = -5\/2<\/p>\n<p>On substituting the values of h and k in equation (1), we get<\/p>\n<p>(2 \u2013 7\/2)<sup>2<\/sup>&nbsp;+ (3 + 5\/2)<sup>2<\/sup>&nbsp;= r<sup>2<\/sup><\/p>\n<p>[(4-7)\/2]<sup>2<\/sup>&nbsp;+ [(6+5)\/2]<sup>2<\/sup>&nbsp;= r<sup>2<\/sup><\/p>\n<p>(-3\/2)<sup>2<\/sup>&nbsp;+ (11\/2)<sup>2<\/sup>&nbsp;= r<sup>2<\/sup><\/p>\n<p>9\/4 + 121\/4 = r<sup>2<\/sup><\/p>\n<p>130\/4 = r<sup>2<\/sup><\/p>\n<p>The equation of the required circle is<\/p>\n<p>(x \u2013 7\/2)<sup>2<\/sup>&nbsp;+ (y + 5\/2)<sup>2<\/sup>&nbsp;= 130\/4<\/p>\n<p>[(2x-7)\/2]<sup>2<\/sup>&nbsp;+ [(2y+5)\/2]<sup>2<\/sup>&nbsp;= 130\/4<\/p>\n<p>4x<sup>2<\/sup>&nbsp;-28x + 49 +4y<sup>2<\/sup>&nbsp;+ 20y + 25 =130<\/p>\n<p>4x<sup>2<\/sup>&nbsp;+4y<sup>2<\/sup>&nbsp;-28x + 20y \u2013 56 = 0<\/p>\n<p>4(x<sup>2<\/sup>&nbsp;+y<sup>2<\/sup>&nbsp;-7x + 5y \u2013 14) = 0<\/p>\n<p>x<sup>2&nbsp;<\/sup>+ y<sup>2&nbsp;<\/sup>\u2013 7x + 5y \u2013 14 = 0<\/p>\n<p>\u2234 The equation of the required circle is x<sup>2&nbsp;<\/sup>+ y<sup>2&nbsp;<\/sup>\u2013 7x + 5y \u2013 14 = 0<\/p>\n<p><strong>12. Find the equation of the circle with radius 5 whose center lies on the x-axis and passes through the point (2, 3).<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let us consider the equation of the required circle be (x \u2013 h)<sup>2<\/sup>+ (y \u2013 k)<sup>2<\/sup>&nbsp;= r<sup>2<\/sup><\/p>\n<p>We know that the radius of the circle is 5 and its center lies on the x-axis, k = 0 and r = 5.<\/p>\n<p>So now, the equation of the circle is (x \u2013 h)<sup>2<\/sup>&nbsp;+ y<sup>2<\/sup>&nbsp;= 25.<\/p>\n<p>It is given that the circle passes through the point (2, 3), so the point will satisfy the equation of the circle.<\/p>\n<p>(2 \u2013 h)<sup>2<\/sup>+ 3<sup>2<\/sup>&nbsp;= 25<\/p>\n<p>(2 \u2013 h)<sup>2<\/sup>&nbsp;= 25-9<\/p>\n<p>(2 \u2013 h)<sup>2<\/sup>&nbsp;= 16<\/p>\n<p>2 \u2013 h =&nbsp;<strong>\u00b1<\/strong>&nbsp;<strong>\u221a<\/strong>16 =&nbsp;<strong>\u00b1<\/strong>&nbsp;4<\/p>\n<p>If 2-h = 4, then h = -2<\/p>\n<p>If 2-h = -4, then h = 6<\/p>\n<p>Then, when h = -2, the equation of the circle becomes<\/p>\n<p>(x + 2)<sup>2<\/sup>&nbsp;+ y<sup>2<\/sup>&nbsp;= 25<\/p>\n<p>x<sup>2<\/sup>&nbsp;+ 12x + 36 + y<sup>2<\/sup>&nbsp;= 25<\/p>\n<p>x<sup>2<\/sup>&nbsp;+ y<sup>2<\/sup>&nbsp;+ 4x \u2013 21 = 0<\/p>\n<p>When h = 6, the equation of the circle becomes<\/p>\n<p>(x \u2013 6)<sup>2<\/sup>&nbsp;+ y<sup>2<\/sup>&nbsp;= 25<\/p>\n<p>x<sup>2<\/sup>&nbsp;-12x + 36 + y<sup>2<\/sup>&nbsp;= 25<\/p>\n<p>x<sup>2<\/sup>&nbsp;+ y<sup>2<\/sup>&nbsp;-12x + 11 = 0<\/p>\n<p>\u2234 The equation of the required circle is x<sup>2<\/sup>&nbsp;+ y<sup>2<\/sup>&nbsp;+ 4x \u2013 21 = 0 and x<sup>2<\/sup>&nbsp;+ y<sup>2<\/sup>&nbsp;-12x + 11 = 0<\/p>\n<p><strong>13. Find the equation of the circle passing through (0,0) and making intercepts a and b on the coordinate axes.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let us consider the equation of the required circle be (x \u2013 h)<sup>2<\/sup>+ (y \u2013 k)<sup>2<\/sup>&nbsp;=r<sup>2<\/sup><\/p>\n<p>We know that the circle passes through (0, 0).<\/p>\n<p>So, (0 \u2013 h)<sup>2<\/sup>+ (0 \u2013 k)<sup>2<\/sup>&nbsp;= r<sup>2<\/sup><\/p>\n<p>h<sup>2<\/sup>&nbsp;+ k<sup>2<\/sup>&nbsp;=&nbsp;r<sup>2<\/sup><\/p>\n<p>Now, the equation of the circle is (x \u2013 h)<sup>2&nbsp;<\/sup>+ (y \u2013 k)<sup>2<\/sup>&nbsp;= h<sup>2<\/sup>&nbsp;+ k<sup>2<\/sup>.<\/p>\n<p>It is given that the circle intercepts a and b on the coordinate axes.<\/p>\n<p>i.e., the circle passes through points (a, 0) and (0, b).<\/p>\n<p>So, (a \u2013 h)<sup>2<\/sup>+ (0 \u2013 k)<sup>2<\/sup>&nbsp;=h<sup>2<\/sup>&nbsp;+k<sup>2<\/sup>\u2026\u2026\u2026\u2026\u2026..(1)<\/p>\n<p>(0 \u2013 h)<sup>2<\/sup>+ (b\u2013 k)<sup>2<\/sup>&nbsp;=h<sup>2<\/sup>&nbsp;+k<sup>2<\/sup>\u2026\u2026\u2026\u2026\u2026\u2026(2)<\/p>\n<p>From equation (1), we obtain<\/p>\n<p>a<sup>2<\/sup>&nbsp;\u2013 2ah + h<sup>2<\/sup>&nbsp;+k<sup>2<\/sup>&nbsp;= h<sup>2<\/sup>&nbsp;+k<sup>2<\/sup><\/p>\n<p>a<sup>2<\/sup>&nbsp;\u2013 2ah = 0<\/p>\n<p>a(a \u2013 2h) =0<\/p>\n<p>a = 0 or (a -2h) = 0<\/p>\n<p>However, a \u2260&nbsp;0; hence, (a -2h) = 0<\/p>\n<p>h = a\/2<\/p>\n<p>From equation (2), we obtain<\/p>\n<p>h<sup>2<\/sup>&nbsp;\u2013 2bk + k<sup>2<\/sup>&nbsp;+ b<sup>2<\/sup>= h<sup>2<\/sup>&nbsp;+k<sup>2<\/sup><\/p>\n<p>b<sup>2<\/sup>&nbsp;\u2013 2bk = 0<\/p>\n<p>b(b\u2013 2k) = 0<\/p>\n<p>b= 0 or (b-2k) =0<\/p>\n<p>However, a&nbsp;\u2260 0; hence, (b -2k) = 0<\/p>\n<p>k&nbsp;=b\/2<\/p>\n<p>So, the equation is<\/p>\n<p>(x \u2013 a\/2)<sup>2<\/sup>&nbsp;+ (y \u2013 b\/2)<sup>2<\/sup>&nbsp;= (a\/2)<sup>2<\/sup>&nbsp;+ (b\/2)<sup>2<\/sup><\/p>\n<p>[(2x-a)\/2]<sup>2<\/sup>&nbsp;+ [(2y-b)\/2]<sup>2<\/sup>&nbsp;= (a<sup>2<\/sup>&nbsp;+ b<sup>2<\/sup>)\/4<\/p>\n<p>4x<sup>2<\/sup>&nbsp;\u2013 4ax + a<sup>2<\/sup>&nbsp;+4y<sup>2<\/sup>&nbsp;\u2013 4by + b<sup>2<\/sup>&nbsp;= a<sup>2<\/sup>&nbsp;+ b<sup>2<\/sup><\/p>\n<p>4x<sup>2<\/sup>&nbsp;+ 4y<sup>2<\/sup>&nbsp;-4ax \u2013 4by = 0<\/p>\n<p>4(x<sup>2<\/sup>&nbsp;+y<sup>2<\/sup>&nbsp;-7x + 5y \u2013 14) = 0<\/p>\n<p>x<sup>2&nbsp;<\/sup>+ y<sup>2&nbsp;<\/sup>\u2013 ax&nbsp;\u2013 by = 0<\/p>\n<p>\u2234 The equation of the required circle is x<sup>2&nbsp;<\/sup>+ y<sup>2&nbsp;<\/sup>\u2013 ax \u2013 by = 0<\/p>\n<p><strong>14. Find the equation of a circle with center (2,2) and passes through the point (4,5).<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>The center of the circle is given as (h, k) = (2,2).<\/p>\n<p>We know that the circle passes through the point (4,5), and the radius (r) of the circle is the distance between points (2,2) and (4,5).<\/p>\n<p>r =&nbsp;<strong>\u221a<\/strong>[(2-4)<sup>2<\/sup>&nbsp;+ (2-5)<sup>2<\/sup>]<\/p>\n<p>= \u221a[(-2)<sup>2<\/sup>&nbsp;+ (-3)<sup>2<\/sup>]<\/p>\n<p>= \u221a[4+9]<\/p>\n<p>= \u221a13<\/p>\n<p>The equation of the circle is given as<\/p>\n<p>(x\u2013 h)<sup>2<\/sup>+ (y \u2013 k)<sup>2<\/sup>&nbsp;= r<sup>2<\/sup><\/p>\n<p>(x \u2013h)<sup>2<\/sup>&nbsp;+ (y \u2013 k)<sup>2<\/sup>&nbsp;= (\u221a13)<sup>2<\/sup><\/p>\n<p>(x \u20132)<sup>2<\/sup>&nbsp;+ (y \u2013 2)<sup>2<\/sup>&nbsp;= (\u221a13)<sup>2<\/sup><\/p>\n<p>x<sup>2<\/sup>&nbsp;\u2013 4x + 4 + y<sup>2&nbsp;<\/sup>\u2013 4y + 4 = 13<\/p>\n<p>x<sup>2<\/sup>&nbsp;+ y<sup>2<\/sup>&nbsp;\u2013 4x \u2013 4y = 5<\/p>\n<p>\u2234 The equation of the required circle is x<sup>2<\/sup>&nbsp;+ y<sup>2<\/sup>&nbsp;\u2013 4x \u2013 4y = 5<\/p>\n<p><strong>15. Does the point (\u20132.5, 3.5) lie inside, outside or on the circle x<sup>2<\/sup>&nbsp;+ y<sup>2<\/sup>&nbsp;= 25?<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>The equation of the given circle is&nbsp;x<sup>2<\/sup>&nbsp;+y<sup>2<\/sup>&nbsp;= 25<\/p>\n<p>x<sup>2<\/sup>&nbsp;+ y<sup>2<\/sup>&nbsp;= 25<\/p>\n<p>(x \u2013 0)<sup>2<\/sup>&nbsp;+ (y \u2013 0)<sup>2<\/sup>&nbsp;= 5<sup>2<\/sup>&nbsp;[which is in the form (x \u2013 h)<sup>2&nbsp;<\/sup>+ (y \u2013 k)<sup>2<\/sup>&nbsp;= r<sup>2<\/sup>]<\/p>\n<p>Where, h = 0, k = 0 and r = 5<\/p>\n<p>So, the distance between the point (-2.5, 3.5) and the center (0,0) is<\/p>\n<p>= \u221a[(-2.5 \u2013 0)<sup>2<\/sup>&nbsp;+ (-3.5 \u2013 0)<sup>2<\/sup>]<\/p>\n<p>= \u221a(6.25 + 12.25)<\/p>\n<p>= \u221a18.5<\/p>\n<p>= 4.3 [which is &lt; 5]<\/p>\n<p>Since the distance between the point (-2.5, -3.5) and the center (0, 0) of the circle is less than the radius of the circle, the point (-2.5, -3.5) lies inside the circle.<\/p>\n<p>We have covered everything related to &#8221; <a href=\"https:\/\/ncert.nic.in\/\" target=\"_blank\" rel=\"noopener\"><strong>NCERT<\/strong><\/a> Solutions Class 11 Maths Chapter 11 Exercise 11.1&#8243;. If you have any doubts, feel free to ask any queries in the comment section below.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-ncert-solutions-for-class-11-maths-chapter-11-exercise-111\"><\/span><strong>FAQs on NCERT Solutions for Class 11 Maths Chapter 11 Exercise 11.1<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>Here are the most frequently asked questions related to NCERT Solutions Class 11 Maths Chapter 11 Exercise 11.1 by the students.<\/p>\n\n\n<div id=\"rank-math-faq\" class=\"rank-math-block\">\n<div class=\"rank-math-list \">\n<div id=\"faq-question-1626858151678\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"what-is-chapter-11-of-ncert-maths-of-class-11\"><\/span><strong>What is Chapter 11 of NCERT Maths of Class 11?<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Chapter 11 of NCERT Maths of Class 11 is Conic Sections.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1626858164560\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"where-can-i-get-class-11-maths-ncert-solutions-for-chapter-11-exercise-111\"><\/span><strong>Where can I get Class 11 Maths NCERT Solutions for Chapter 11 Exercise 11.1?<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can get the NCERT Solutions for Class 11 Maths Chapter 11 Exercise 11.1 from the pdf given in the article above.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1626858183490\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"how-many-exercises-are-there-in-chapter-11-of-maths-ncert-of-class-11\"><\/span><strong>How many exercises are there in Chapter 11 of Maths NCERT of Class 11?<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>There are a total of 4 exercises and one miscellaneous exercise in chapter 11 of Maths NCERT of Class 11.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1626858211840\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"what-are-some-of-the-benefits-of-using-ncert-math-solutions\"><\/span><strong>What are some of the benefits of using NCERT Math Solutions?<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Maths NCERT Solutions has the advantage of allowing you to learn the proper solution while studying the problem and using it to guide you in the right direction. You will be able to save time as a result of it.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1626858240545\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"what-is-exercise-111-of-the-ncert-maths-of-class-11-chapter-11-based-on\"><\/span><strong>What is exercise 11.1 of the NCERT Maths of Class 11 Chapter 11 based on?<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>The following subjects are covered in Exercise 11.1 of NCERT Solutions for Class 11 Maths Chapter 11- Conic Sections:<br \/>Introduction<br \/>Sections of a Cone<br \/>Circle, ellipse, parabola and hyperbola<br \/>Degenerated conic sections<br \/>Circle<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1626858253794\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"what-are-ncert-maths-solutions-for-class-11\"><\/span><strong>What are NCERT Maths Solutions for Class 11?<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Answers have been produced by NCERT specialists based on the most recent CBSE guidelines. All chapters from the CBSE 11th-grade math textbooks are included.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions for Class 11 Maths Chapter 11 Exercise 11.1: All exercise 11.1 questions are answered in the NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections. The goal of practicing NCERT Solutions is to improve your final test result. To obtain the solutions to all of the exercises in all of the chapters, &#8230; <a title=\"NCERT Solutions Class 11 Maths Chapter 11 Exercise 11.1\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/ncert-solutions-for-class-11-maths-chapter-11-exercise-11-1\/\" aria-label=\"More on NCERT Solutions Class 11 Maths Chapter 11 Exercise 11.1\">Read more<\/a><\/p>\n","protected":false},"author":248,"featured_media":146042,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73413,2934,2985,73410],"tags":[76386,3570,76234],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/106669"}],"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\/248"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=106669"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/106669\/revisions"}],"predecessor-version":[{"id":574313,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/106669\/revisions\/574313"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/146042"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=106669"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=106669"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=106669"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}