{"id":6779,"date":"2021-10-21T16:25:54","date_gmt":"2021-10-21T10:55:54","guid":{"rendered":"https:\/\/mathemerize.com\/?p=6779"},"modified":"2021-10-23T16:00:13","modified_gmt":"2021-10-23T10:30:13","slug":"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","status":"publish","type":"post","link":"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\/","title":{"rendered":"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, is"},"content":{"rendered":"

Solution :<\/h2>\n

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

Similar Questions<\/h3>\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 \u2013 4) + y(y \u2013 3) = 0.<\/a><\/p>\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","protected":false},"excerpt":{"rendered":"

Solution : Family of circles is \\(x^2 + y^2 – 2x – 4y + 1\\) + \\(\\lambda\\)(\\(x^2 + y^2 – 1\\)) = 0 (1 + \\(\\lambda\\))\\(x^2\\) + (1 + \\(\\lambda\\))\\(y^2\\) – 2x – 4y + (1 – \\(\\lambda\\))) = 0 \\(x^2 + y^2 – {2\\over {1 + \\lambda}} x – {4\\over {1 + \\lambda}}y + …<\/p>\n

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, is<\/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":"\nThe equation of the circle through the points of intersection of \\(x^2 + y^2 - 1\\) = 0, \\(x^2 + y^2 - 2x - 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