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A triangle has vertices A(0,0), B(8,0) and C(2,6). Its nine-point…
Question
A triangle has vertices A(0,0), B(8,0) and C(2,6). Its nine-point circle (the circle through the three midpoints of the sides, the three feet of the altitudes, and the three Euler points) has radius equal to half the circumradius. Find the square of the radius of the nine-point circle.
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Step-by-step solution
The nine-point circle radius is R/2 where R is the circumradius, so its squared radius is R^2/4.
Find the circumcentre O = (x,y). It is equidistant from A(0,0) and B(8,0): x^2 + y^2 = (x-8)^2 + y^2 => 0 = -16x + 64 => x = 4.
Equidistant from A and C(2,6): x^2 + y^2 = (x-2)^2 + (y-6)^2 => 0 = -4x + 4 - 12y + 36 => 4x + 12y = 40 => with x = 4: 16 + 12y = 40 => y = 2.
So O = (4,2).
R^2 = OA^2 = 4^2 + 2^2 = 20.
Nine-point circle radius^2 = R^2/4 = 20/4 = 5.
Final answer5
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