{"id":3875,"date":"2021-08-09T19:53:04","date_gmt":"2021-08-09T19:53:04","guid":{"rendered":"https:\/\/mathemerize.com\/?p=3875"},"modified":"2021-11-20T16:26:47","modified_gmt":"2021-11-20T10:56:47","slug":"family-of-circles","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/family-of-circles\/","title":{"rendered":"Family of Circles – Equations & Examples"},"content":{"rendered":"
Here you will learn equation of family of circles with examples.<\/p>\n
Let’ begin –<\/p>\n
(a)<\/strong>\u00a0 The equation of the family of circles passing through the point of intersection of two circles \\(S_1\\) = 0 & \\(S_2\\) = 0 is :<\/p>\n \\(S_1\\) + K\\(S_2\\) = 0\u00a0 \u00a0 \u00a0 (K \\(\\ne\\) -1)<\/strong><\/p>\n<\/blockquote>\n (b)<\/strong>\u00a0 The equation of the family of circles passing through the point of intersection of circle S = 0 & a line L = 0 is given by<\/p>\n S + KL = 0.<\/strong><\/p>\n<\/blockquote>\n (c)\u00a0<\/strong> The equation of the family of circles passing through two given points (\\(x_1, y_1\\)) & (\\(x_2, y_2\\)) can be written in the form :<\/p>\n\n\n (x – \\(x_1\\))(x – \\(x_2\\)) + (y – \\(y_1\\))(y – \\(y_2\\)) + K\\begin{vmatrix}\n\t\t x & y & 1 \\\\ \n\t\t\tx_1 & y_1 & 1 \\\\\n\t\t\tx_2 & y_2 & 1\n\t\t\t\\end{vmatrix}<\/span> = 0 where K is a parameter. (d)\u00a0<\/strong> The equation of the family of circles touching a fixed line y – \\(y_1\\) = m(x – \\(x_1\\)) at the fixed point (\\(x_1, y_1\\)) is \\({(x – x_1)}^2\\) + \\({(y – y_1)}^2\\) + K [y – \\(y_1\\) – m(x – \\(x_1\\))] = 0,where K is a parameter.<\/p>\n (e)<\/strong>\u00a0 Family of circles circumscribing a triangle whose sides are given by \\(L_1\\) = 0 ; \\(L_2\\) = 0 & \\(L_3\\) = 0 is given by : \\(L_1\\)\\(L_2\\) + \\(\\lambda\\)\\(L_2\\)\\(L_3\\) + \\(\\mu\\)\u00a0 \\(L_3\\)\\(L_1\\) = 0 provided coefficients of xy = 0 & coefficient of \\(x^2\\) = coefficient of \\(y^2\\).<\/p>\n (f)<\/strong>\u00a0 Family of circles circumscribing a quadrilateral whose sides are given by \\(L_1\\) = 0, \\(L_2\\) = 0, \\(L_3\\) = 0 & \\(L_4\\) = 0 is \\(L_1\\)\\(L_3\\) + \\(\\lambda\\)\\(L_2\\)\\(L_4\\) provided coefficients of xy = 0 & coefficient of \\(x^2\\) = coefficient of \\(y^2\\).<\/p>\n Example<\/span><\/strong> : Find the equation of the circle through the points of intersection of \\(x^2 + y^2 – 1\\) = 0, \\(x^2 + y^2 – 2x – 4y + 1\\) = 0 and touching the line x + 2y = 0.<\/p>\n Solution<\/strong><\/span> : Family of circles is \\(x^2 + y^2 – 2x – 4y + 1\\) + \\(\\lambda\\)(\\(x^2 + y^2 – 1\\)) = 0<\/p>\n (1 + \\(\\lambda\\))\\(x^2\\) + (1 + \\(\\lambda\\))\\(y^2\\) – 2x – 4y + (1 – \\(\\lambda\\))) = 0<\/p>\n \\(x^2 + y^2 – {2\\over {1 + \\lambda}} x – {4\\over {1 + \\lambda}}y + {{1 – \\lambda}\\over {1 + \\lambda}}\\) = 0<\/p>\n Centre is (\\({1\\over {1 + \\lambda}}\\), \\({2\\over {1 + \\lambda}}\\))\u00a0 and radius = \\(\\sqrt{4 + {\\lambda}^2}\\over |1 + \\lambda|\\)<\/p>\n Since it touches the line x + 2y = 0, hence<\/p>\n Radius = Perpendicular distance from center to the line<\/p>\n i.e., |\\({1\\over {1 + \\lambda}} + 2{2\\over {1 + \\lambda}}\\over \\sqrt{1^2 + 2^2}\\)| = \\(\\sqrt{4 + {\\lambda}^2}\\over |1 + \\lambda|\\) \\(\\implies\\) \\(\\sqrt{5}\\) = \\(\\sqrt{4 + {\\lambda}^2}\\) \\(\\implies\\) \\(\\lambda\\) = \\(\\pm\\) 1.<\/p>\n \\(\\lambda\\) = -1 cannot be possible in case of circle. So \\(\\lambda\\) = 1.<\/p>\n Hence the equation of the circle is \\(x^2 + y^2 – x – 2y\\) = 0<\/p>\n\n\n\n
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