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
x^{2} + y^{2} = 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, 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.
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, 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.
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
x^{2} + (-1/3)^{2} = 1
x^{2 }+ 1/9 = 1
Subtract 1/9 from each side.
x^{2} = 1 - 1/9
x^{2} = 9/9 - 1/9
x^{2} = (9 - 1)/9
x^{2} = 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} + k^{2} = 1
3/4 + k^{2} = 1
Subtract 3/4 from each side.
k^{2} = 1 - 3/4
k^{2} = 4/4 - 3/4
k^{2} = (4 - 3)/4
k^{2} = 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} + y^{2} = 1
4/25 + y^{2} = 1
Subtract 4/25 from each side.
y^{2} = 1 - 4/25
y^{2} = 25/25 - 4/25
y^{2} = (25 - 4)/25
y^{2} = 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
x^{2} + (-1/2)^{2} = 1
x^{2 }+ 1/4 = 1
Subtract 1/4 from each side.
x^{2} = 1 - 1/4
x^{2} = 4/4 - 1/4
x^{2} = (4 - 1)/4
x^{2} = 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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