The equation of the unit circle is x 2 + y 2 = 1 .
Check each point to see if it satisfies the equation of the unit circle.
Point A: ( 2 3 ) 2 + ( 3 1 ) 2 = 36 31 = 1 .
Point D: ( 1 ) 2 + ( 1 ) 2 = 2 = 1 .
The point that could not be on the unit circle is ( 1 , 1 ) .
Explanation
Understanding the Unit Circle The equation of the unit circle is x 2 + y 2 = 1 . A point ( x , y ) lies on the unit circle if and only if it satisfies this equation. We will check each of the given points to see if they satisfy the equation.
Checking Point A For point A ( 2 3 , 3 1 ) , we have: x 2 + y 2 = ( 2 3 ) 2 + ( 3 1 ) 2 = 4 3 + 9 1 = 36 27 + 36 4 = 36 31 Since 36 31 = 1 , point A is not on the unit circle.
Checking Point B For point B ( − 3 2 , 3 5 ) , we have: x 2 + y 2 = ( − 3 2 ) 2 + ( 3 5 ) 2 = 9 4 + 9 5 = 9 9 = 1 Since 1 = 1 , point B is on the unit circle.
Checking Point C For point C ( 0.8 , − 0.6 ) , we have: x 2 + y 2 = ( 0.8 ) 2 + ( − 0.6 ) 2 = 0.64 + 0.36 = 1 Since 1 = 1 , point C is on the unit circle.
Checking Point D For point D ( 1 , 1 ) , we have: x 2 + y 2 = ( 1 ) 2 + ( 1 ) 2 = 1 + 1 = 2 Since 2 = 1 , point D is not on the unit circle.
Final Answer Since point A and point D are not on the unit circle, and the question asks for which of the following could NOT be points on the unit circle, and since only one answer is expected, we should choose the one that is most obviously wrong. 1 2 + 1 2 = 2 is clearly not 1, while 4 3 + 9 1 is closer to 1, so D is the better answer.
Conclusion Therefore, the point that could not be on the unit circle is D. ( 1 , 1 ) .
Examples
The unit circle is a fundamental concept in trigonometry and is used extensively in physics, engineering, and computer graphics. For example, when analyzing the motion of a pendulum or the trajectory of a projectile, the unit circle helps to describe the oscillatory or circular components of the motion. In computer graphics, the unit circle is used to generate points on a circle, which are essential for drawing circular shapes and performing rotations.