{"id":6776,"date":"2021-10-21T15:55:10","date_gmt":"2021-10-21T10:25:10","guid":{"rendered":"https:\/\/mathemerize.com\/?p=6776"},"modified":"2021-10-23T16:02:53","modified_gmt":"2021-10-23T10:32:53","slug":"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","status":"publish","type":"post","link":"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\/","title":{"rendered":"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 – 4) + y(y – 3) = 0."},"content":{"rendered":"

Solution :<\/h2>\n

Pair of normals are (x + 2y)(x + 3) = 0<\/p>\n

\\(\\therefore\\) Normals are x + 2y = 0, x + 3 = 0<\/p>\n

Point of intersection of normals is the center of the required circle i.e. \\(C_1\\)(-3,3\/2) and center of the given circle is<\/p>\n

\\(C_2\\)(2,3\/2) and radius \\(r^2\\) = \\(\\sqrt{4 + {9\\over 4}}\\) = \\(5\\over 2\\)<\/p>\n

Let \\(r_1\\) be the radius of the required circle<\/p>\n

\\(\\implies\\)\u00a0 \\(r_1\\) = \\(C_1\\)\\(C_2\\) + \\(r_2\\) = \\(\\sqrt{(-3-2)^2 + ({3\\over 2}- {3\\over 2})^2}\\) + \\(5\\over 2\\) = \\(15\\over 2\\)<\/p>\n

Hence equation of required circle is \\(x^2 + y^2 + 6x – 3y – 45\\) = 0<\/p>\n

\n

Similar Questions<\/h3>\n

Find the equation of the normal to the circle \\(x^2 + y^2 \u2013 5x + 2y -48\\) = 0 at the point (5,6).<\/a><\/p>\n

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

The length of the diameter of the circle which touches the X-axis at the point (1,0) and passes through the point (2,3) is<\/a><\/p>\n

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

The circle passing through (1,-2) and touching the axis of x at (3, 0) also passes through the point<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"

Solution : Pair of normals are (x + 2y)(x + 3) = 0 \\(\\therefore\\) Normals are x + 2y = 0, x + 3 = 0 Point of intersection of normals is the center of the required circle i.e. \\(C_1\\)(-3,3\/2) and center of the given circle is \\(C_2\\)(2,3\/2) and radius \\(r^2\\) = \\(\\sqrt{4 + {9\\over …<\/p>\n

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 – 4) + y(y – 3) = 0.<\/span> Read More »<\/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":"\nFind 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 - 4) + y(y - 3) = 0.<\/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-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\/\" \/>\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 circle having the lines \\(x^2\\) + 2xy + 3x + 6y = 0 as its normal and having size just sufficient to contain the circle x(x - 4) + y(y - 3) = 0.\" \/>\n<meta property=\"og:description\" content=\"Solution : Pair of normals are (x + 2y)(x + 3) = 0 (therefore) Normals are x + 2y = 0, x + 3 = 0 Point of intersection of normals is the center of the required circle i.e. (C_1)(-3,3\/2) and center of the given circle is (C_2)(2,3\/2) and radius (r^2) = (sqrt{4 + {9over … 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 – 4) + y(y – 3) = 0. 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