{"id":6773,"date":"2021-10-21T15:34:36","date_gmt":"2021-10-21T10:04:36","guid":{"rendered":"https:\/\/mathemerize.com\/?p=6773"},"modified":"2021-10-23T16:12:41","modified_gmt":"2021-10-23T10:42:41","slug":"find-the-equation-of-the-normal-to-the-circle-x2-y2-5x-2y-48-0-at-the-point-56","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/find-the-equation-of-the-normal-to-the-circle-x2-y2-5x-2y-48-0-at-the-point-56\/","title":{"rendered":"Find the equation of the normal to the circle \\(x^2 + y^2 &#8211; 5x + 2y -48\\) = 0 at the point (5,6)."},"content":{"rendered":"<h2>Solution :<\/h2>\n<p>Since the normal to the circle always passes through the center,<\/p>\n<p>so the equation of the normal will be the line passing through (5,6) &amp; (\\(5\\over 2\\), -1)<\/p>\n<p>i.e.\u00a0 y + 1 = \\(7\\over {5\/2}\\)(x &#8211; \\(5\\over 2\\)) \\(\\implies\\) 5y + 5 = 14x &#8211; 35<\/p>\n<p>\\(\\implies\\)\u00a0 14x &#8211; 5y &#8211; 40 = 0<\/p>\n<hr \/>\n<h3 style=\"text-align: center;\">Similar Questions<\/h3>\n<p><a class=\"row-title\" href=\"https:\/\/mathemerize.com\/if-the-lines-3x-4y-70-and-2x-3y-50-are-two-diameters-of-a-circle-of-area-49%cf%80-square-units-then-what-is-the-equation-of-the-circle\/\" aria-label=\"\u201cIf the lines 3x-4y-7=0 and 2x-3y-5=0 are two diameters of a circle of area 49\u03c0 square units, then what is the equation of the circle?\u201d (Edit)\">If the lines 3x-4y-7=0 and 2x-3y-5=0 are two diameters of a circle of area 49\u03c0 square units, then what is the equation of the circle?<\/a><\/p>\n<p><a class=\"row-title\" href=\"https:\/\/mathemerize.com\/find-the-equation-of-circle-having-the-lines-x2-2xy-3x-6y-0-as-its-normal-and-having-size-just-sufficient-to-contain-the-circle-xx-4-yy-3-0\/\" aria-label=\"\u201cFind the equation of circle having the lines \\(x^2\\) + 2xy + 3x + 6y = 0 as its normal and having size just sufficient to contain the circle x(x \u2013 4) + y(y \u2013 3) = 0.\u201d (Edit)\">Find the equation of circle having the lines \\(x^2\\) + 2xy + 3x + 6y = 0 as its normal and having size just sufficient to contain the circle x(x \u2013 4) + y(y \u2013 3) = 0.<\/a><\/p>\n<p><a class=\"row-title\" href=\"https:\/\/mathemerize.com\/the-equation-of-the-circle-through-the-points-of-intersection-of-x2-y2-1-0-x2-y2-2x-4y-1-0-and-touching-the-line-x-2y-0-is\/\" aria-label=\"\u201cThe equation of the circle through the points of intersection of \\(x^2 + y^2 \u2013 1\\) = 0, \\(x^2 + y^2 \u2013 2x \u2013 4y + 1\\) = 0 and touching the line x + 2y = 0, is\u201d (Edit)\">The equation of the circle through the points of intersection of \\(x^2 + y^2 \u2013 1\\) = 0, \\(x^2 + y^2 \u2013 2x \u2013 4y + 1\\) = 0 and touching the line x + 2y = 0, is<\/a><\/p>\n<p><a class=\"row-title\" href=\"https:\/\/mathemerize.com\/the-circle-passing-through-1-2-and-touching-the-axis-of-x-at-3-0-also-passes-through-the-point\/\" aria-label=\"\u201cThe circle passing through (1,-2) and touching the axis of x at (3, 0) also passes through the point\u201d (Edit)\">The circle passing through (1,-2) and touching the axis of x at (3, 0) also passes through the point<\/a><\/p>\n<p><a class=\"row-title\" href=\"https:\/\/mathemerize.com\/the-equation-of-the-circle-passing-through-the-foci-of-the-ellipse-x2over-16-y2over-9-1-and-having-center-at-0-3-is\/\" aria-label=\"\u201cThe equation of the circle passing through the foci of the ellipse \\(x^2\\over 16\\) + \\(y^2\\over 9\\) = 1 and having center at (0, 3) is\u201d (Edit)\">The equation of the circle passing through the foci of the ellipse \\(x^2\\over 16\\) + \\(y^2\\over 9\\) = 1 and having center at (0, 3) is<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Solution : Since the normal to the circle always passes through the center, so the equation of the normal will be the line passing through (5,6) &amp; (\\(5\\over 2\\), -1) i.e.\u00a0 y + 1 = \\(7\\over {5\/2}\\)(x &#8211; \\(5\\over 2\\)) \\(\\implies\\) 5y + 5 = 14x &#8211; 35 \\(\\implies\\)\u00a0 14x &#8211; 5y &#8211; 40 = &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"\" href=\"https:\/\/mathemerize.com\/find-the-equation-of-the-normal-to-the-circle-x2-y2-5x-2y-48-0-at-the-point-56\/\"> <span class=\"screen-reader-text\">Find the equation of the normal to the circle \\(x^2 + y^2 &#8211; 5x + 2y -48\\) = 0 at the point (5,6).<\/span> Read More &raquo;<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-global-header-display":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":""},"categories":[45,43],"tags":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v20.12 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Find the equation of the normal to the circle \\(x^2 + y^2 - 5x + 2y -48\\) = 0 at the point (5,6).<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mathemerize.com\/find-the-equation-of-the-normal-to-the-circle-x2-y2-5x-2y-48-0-at-the-point-56\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Find the equation of the normal to the circle \\(x^2 + y^2 - 5x + 2y -48\\) = 0 at the point (5,6).\" \/>\n<meta property=\"og:description\" content=\"Solution : Since the normal to the circle always passes through the center, so the equation of the normal will be the line passing through (5,6) &amp; ((5over 2), -1) i.e.\u00a0 y + 1 = (7over {5\/2})(x &#8211; (5over 2)) (implies) 5y + 5 = 14x &#8211; 35 (implies)\u00a0 14x &#8211; 5y &#8211; 40 = &hellip; Find the equation of the normal to the circle (x^2 + y^2 &#8211; 5x + 2y -48) = 0 at the point (5,6). 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