How do you find six sets of points in the first quadrant of the unit circle with a radius of 3/2

Question

List 6 points (x,y) that lie on the unit circle (circle with center at the origin and radius 1) and also are in the first quadrant. Plot the points. Use your answer from the first question to find 6 points in the first quadrant that lie on the circle centered at the origin with a radius 3/2. The points (3,4) are not on the unit circle. What scalar multiples of (3,4) are on the unit circle? 

Answer 1

The equation of a circle at the origin is x² + y² = r² where r represents the radius of the circle. In the case of a circle with radius 1 centered at the origin, the equation would be:

 

x² + y² = 1

 

Points in the first quadrant that satisfy this equation must have positive abscissas and ordinates. You can find these by putting a random abscissa into the equation & solve for the ordinate. So I chose 3 random values for the abscissas whose value was smaller than that of the radius: 1/2, 3/5, 2/3. I put these into the equation & found the corresponding ordinates to make 3 coordinates that satisfy the equation. Since the commutative property works for this equation, the switching of the abscissas and ordinates also created 3 other coordinates for a total of 6 that satisfy this equation. They are:

 

(3/5, 4/5) 

(4/5, 3/5)

(1/2, √3/2)

(√3/2, 1/2)

(2/3, √5/3)

(√5/3, 2/3)

 

For a circle with radius 3/2 centered at the origin, the equation of that circle would be:

 

x² + y² = 9/4 

 

I will use 1/2, 9/10, 1.

 

(1/2, √2)

(√2, 1/2)

(9/10, 6/5)

(6/5, 9/10)

(1, √5/2)

(√5/2, 1)