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

The unit circle is the circle of radius 1 centered on the origin within the xy-plane.

Its equation is

x^{2} + y^{2} = 1

## A Level on the Unit Circle

Instance 1 :

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

Answer :

Now we have to indicate that this level satisfies the equation of the unit circle, that’s, x^{2} + y^{2} = 1.

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

= (16 + 9)/25

= 25/25

= 1

So, A is on the unit circle.

Instance 2 :

Present that the purpose B(√3/3, √6/3) is on the unit circle.

Answer :

Now we have to indicate that this level satisfies the equation of the unit circle, that’s, x^{2} + y^{2} = 1.

(√3/3)^{2} + (√6/3)^{2} = 3/9 + 6/9

= (3 + 6)/9

= 9/9

= 1

So, P is on the unit circle.

## Finding a Level on the Unit Circle

Instance 3 :

The purpose P is on the unit circle. Discover P(x, y) from the given data. The y-coordinate of P is -1/3 and the x-coordinate is constructive.

Answer :

As a result of the purpose is on the unit circle, now we have

x^{2} + (-1/3)^{2} = 1

x^{2 }+ 1/9 = 1

Subtract 1/9 from all sides.

x^{2} = 1 – 1/9

x^{2} = 9/9 – 1/9

x^{2} = (9 – 1)/9

x^{2} = 8/9

Take sq. root on either side.

x = ± 2√2/3

As a result of x-coordinate is detrimental,

x = 2√2/3

The purpose is

P(2√2/3, -1/3)

Instance 4 :

If the the purpose P(√3/2, ok) is on the unit circle in quadrant IV, discover P(x, y).

Answer :

As a result of the purpose is on the unit circle, now we have

(√3/2)^{2} + ok^{2} = 1

3/4 + ok^{2} = 1

Subtract 3/4 from all sides.

ok^{2} = 1 – 3/4

ok^{2} = 4/4 – 3/4

ok^{2} = (4 – 3)/4

ok^{2} = 1/4

Take sq. root on either side.

ok = ± 1/2

As a result of the purpose is in quadrant IV and ok is the y-coordinate, the worth of ok have to be detrimental.

ok = -1/2

The purpose is

P(√3/2, -1/2)

Instance 5 :

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

Answer :

As a result of the purpose is on the unit circle, now we have

(-2/5)^{2} + y^{2} = 1

4/25 + y^{2} = 1

Subtract 4/25 from all sides.

y^{2} = 1 – 4/25

y^{2} = 25/25 – 4/25

y^{2} = (25 – 4)/25

y^{2} = 21/25

Take sq. root on either side.

y = ± √21/5

As a result of P lies above the x-axis, y-coordinate have to be constructive.

y = √21/5

The purpose is

P(-2/5, √21/5)

Instance 6 :

The purpose P is on the unit circle. Discover P(x, y) from the given data. The y-coordinate of P is -1/2 and P lies on the left aspect of y-axis.

Answer :

As a result of the purpose is on the unit circle, now we have

x^{2} + (-1/2)^{2} = 1

x^{2 }+ 1/4 = 1

Subtract 1/4 from all sides.

x^{2} = 1 – 1/4

x^{2} = 4/4 – 1/4

x^{2} = (4 – 1)/4

x^{2} = 3/4

Take sq. root on either side.

x = ± √3/2

As a result of P lies on the left aspect of y-axis, x-coordinate have to be detrimental.

x = –√3/2

The purpose is

P(–√3/2, -1/2)

## Associated Stuff

1. Terminal Factors on the Unit Circle

2. Reference Quantity on the Unit Circle

3. Utilizing Reference Quantity to Discover terminal Factors

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