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The circumcenter is at (137/26, 23/26) = (5 7/26, 23/26)

The circumcenter is where the side perpendicular bisectors meet; this can be calculated by finding the equations of the perpendicular bisectors and solving the simultaneous equations that result (only 2 needs be solved, the third can be used as a check):

Let the three points of the triangle be A (2, 3), B (5, -3), C (9, 2)

The sides are AB, BC, AC with perpendicular bisectors AB', BC', AC' respectively; then:

AB has midpoint ((5+2)/2, (-3 + 3)) = (7/2, 0)

AB has gradient (-3 - 3)/(5 - 2) = -6/3 = -2

→ AB' has gradient 1/2

→ AB' has equation y - 0 = 1/2(x - 7/2) → y = 1/2(x - 7/2)

BC has midpoint ((9 + 5)/2, (2 + -3)/2) = (7, -1/2)

BC has gradient (2 - -3)/(9 - 5) = 5/4

→ BC' has gradient -4/5

→ BC' has equation y - -1/2 = -4/5(x - 7) → y = 4/5(7 - x) - 1/2

AC has midpoint ((9 + 2)/2, (2 + 3)/2) = (11/2, 5/2)

AC has gradient (2 - 3)/(9 - 2) = -1/7

→ AC' has gradient 7

→ AC' has equation y - 5/2 = 7(x - 11/2) → y = 7(x - 11/2) + 5/2

Solving for point of intersection of AB' and AC':

1/2(x - 7/2) = 7(x - 11/2) + 5/2

→ x - 7/2 = 14x - 77 + 5

→ 13x = 72 - 7/2 = 137/2

→ x = 137/26 (= 5 7/26)

Using AB'

y = 1/2(137/26 - 7/2)

→ y = 1/2(137/26 - 91/26)

→ y = 23/26

Checking using BC':

y = 4/5(7 - x) - 1/2

→ y = 4/5(7 - 137/26) - 1/2

→ y = 4/5(182/26-137/26) - 13/26

→ y = 4/5(45/26) - 13/26

→ y = 36/26 - 13/26 = 23/26 as expected.

Q: What is the circumcenter of the triangle with the points (23) (5-3) and (92)?

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