# THE UNIT CIRCLE

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

## A Point on the Unit Circle

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.

## Locating a Point 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

x+ 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

x+ 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)

## Related Stuff

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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