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Find the coordinates of the circum-centre of the triangle whose vertices are (0, 0), (8,0) and (0,6). Find the Circum-radius also.
(4, 3),6
(3,4),5
(4, 3),5
(4, 3),3
Circum-centre is the point of intersection of the perpendicular bisectors of the three sides of a triangle. Let S (x, y) be the circumcentre. SA = SB = SC. ∴ &redic;(x2 + y2) = &redic; (x-8)2 + (y-2) )2. &redic;(x2 + y2) = &redic; (x-0)2 + (y-6) )2. Squaring x2 + y2 = x2 – 16x + 64 + y2. So x = 4. Also x2 + y2 = x2 + y2 – 12y + 36. So y = 3. Hence coordinates of the circumcentre is (4, 3). Circumradius = SA = &redic; x2 + y = & redic;16+9 = 5. => (3,4),5
By: Amit Kumar ProfileResourcesReport error
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