(1/ 2 , 3/ 2) We check the point by showing that the coordinates satisfy the equation of the unit circle. x^2 + y^2 = 1 /2^ 2 + (?)^ 2 = 1 /4 + (?)/ 4 = (?)Thus, the point 1 /2 , square 3 /2 does lie on the graph of the unit circle.
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x²+y²=1 is the equation of the unit circle with centre at the origin.

The point (½,√3/2) lies on the circle because ½²+(√3/2)²=¼+¾=1.
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