The set of points at a distance 1 from the origin is a circle of radius 1.

The unit circle is the circle of radius 1 centered at the origin in the xy-plane.

Its equation is

x2 + y2 = 1

Example 1 :

Show that the point A(4/5, -3/5) is on the unit circle.

Solution :

We have to show that this point satisfies the equation of the unit circle, that is, x2 + y2 = 1.

(4/5)2 + (-3/5)2 = 16/25 + 9/25

= (16 + 9)/25

= 25/25

= 1

So, A is on the unit circle.

Example 2 :

Show that the point B(√3/3, √6/3) is on the unit circle.

Solution :

We have to show that this point satisfies the equation of the unit circle, that is, x2 + y2 = 1.

(√3/3)2 + (√6/3)2 = 3/9 + 6/9

= (3 + 6)/9

= 9/9

= 1

So, P is on the unit circle.

Example 3 :

The point P is on the unit circle. Find P(x, y) from the given information. The y-coordinate of P is -1/3 and the x-coordinate is positive.

Solution :

Because the point is on the unit circle, we have

x2 + (-1/3)2 = 1

x2 + 1/9 = 1

Subtract 1/9 from each side.

x2 = 1 – 1/9

x2 = 9/9 – 1/9

x2 = (9 – 1)/9

x2 = 8/9

Take square root on both sides.

x = ± 2√2/3

Because x-coordinate is negative,

x = 2√2/3

The point is

P(2√2/3, -1/3)

Example 4 :

If the the point P(√3/2, k) is on the unit circle in quadrant IV, find P(x, y).

Solution :

Because the point is on the unit circle, we have

(√3/2)2 + k2 = 1

3/4 + k2 = 1

Subtract 3/4 from each side.

k2 = 1 – 3/4

k2 = 4/4 – 3/4

k2 = (4 – 3)/4

k2 = 1/4

Take square root on both sides.

k = ± 1/2

Because the point is in quadrant IV and k is the y-coordinate, the value of k must be negative.

k = -1/2

The point is

P(√3/2, -1/2)

Example 5 :

The point P is on the unit circle. Find P(x, y) from the given information. The x-coordinate of P is -2/5 and P lies above the x-axis.

Solution :

Because the point is on the unit circle, we have

(-2/5)2 + y2 = 1

4/25 + y2 = 1

Subtract 4/25 from each side.

y2 = 1 – 4/25

y2 = 25/25 – 4/25

y2 = (25 – 4)/25

y2 = 21/25

Take square root on both sides.

y = ± √21/5

Because P lies above the x-axis, y-coordinate must be positive.

y = √21/5

The point is

P(-2/5, √21/5)

Example 6 :

The point P is on the unit circle. Find P(x, y) from the given information. The y-coordinate of P is -1/2 and P lies on the left side of y-axis.

Solution :

Because the point is on the unit circle, we have

x2 + (-1/2)2 = 1

x2 + 1/4 = 1

Subtract 1/4 from each side.

x2 = 1 – 1/4

x2 = 4/4 – 1/4

x2 = (4 – 1)/4

x2 = 3/4

Take square root on both sides.

x = ± √3/2

Because P lies on the left side of y-axis, x-coordinate must be negative.

x = -√3/2

The point is

P(-√3/2, -1/2)

1. Terminal Points on the Unit Circle

2. Reference Number on the Unit Circle

3. Using Reference Number to Find terminal Points

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