{"id":6921,"date":"2021-10-22T01:28:33","date_gmt":"2021-10-21T19:58:33","guid":{"rendered":"https:\/\/mathemerize.com\/?p=6921"},"modified":"2021-10-23T15:19:16","modified_gmt":"2021-10-23T09:49:16","slug":"the-equation-of-the-circle-passing-through-the-foci-of-the-ellipse-x2over-16-y2over-9-1-and-having-center-at-0-3-is","status":"publish","type":"post","link":"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\/","title":{"rendered":"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"},"content":{"rendered":"

Solution :<\/h2>\n

Given the equation of ellipse is \\(x^2\\over 16\\) + \\(y^2\\over 9\\) = 1<\/p>\n

Here, a = 4, b = 3, e = \\(\\sqrt{1-{9\\over 16}}\\) = \\(\\sqrt{7\\over 4}\\)<\/p>\n

\\(\\therefore\\) foci is (\\(\\pm ae\\), 0) = (\\(\\pm\\sqrt{7}\\), 0)<\/p>\n

\\(\\therefore\\) Radius of the circle, r = \\(\\sqrt{(ae)^2+b^2}\\)<\/p>\n

r = \\(\\sqrt{7+9}\\) = \\(\\sqrt{16}\\) = 4<\/p>\n

Now, equation of the circle is \\((x-0)^2 + (y-3)^2\\) = 16<\/p>\n

\\(\\therefore\\) \\(x^2 + y^2 – 6y – 7\\) = 0<\/p>\n

\n

Similar Questions<\/h3>\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

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 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

Let C be the circle with center at (1,1) and radius 1. If T is the circle centered at (0,y) passing through origin and touching the circle C externally, then the radius of T is equal to<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"

Solution : Given the equation of ellipse is \\(x^2\\over 16\\) + \\(y^2\\over 9\\) = 1 Here, a = 4, b = 3, e = \\(\\sqrt{1-{9\\over 16}}\\) = \\(\\sqrt{7\\over 4}\\) \\(\\therefore\\) foci is (\\(\\pm ae\\), 0) = (\\(\\pm\\sqrt{7}\\), 0) \\(\\therefore\\) Radius of the circle, r = \\(\\sqrt{(ae)^2+b^2}\\) r = \\(\\sqrt{7+9}\\) = \\(\\sqrt{16}\\) = 4 Now, equation …<\/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<\/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 passing through the foci of the ellipse \\(x^2\\over 16\\) + \\(y^2\\over 9\\) = 1 and having center 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