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Challenge

Investigating Points on a Circle's Circumference

In the diagram below, a circle centered at the origin with radius

5

has been drawn. Some points that lie on the circle can be identified.

identify points

Given that

P

is a point that lies on the circle and the

x -

coordinate of

P

is

1

, can you determine the

y -

coordinate of

P ?

Discussion

Deriving the Equation of a Circle at the Origin

The equation of a circle on a coordinate plane can be obtained using the Distance Formula. For example, consider a circle centered at the origin with a radius of

2 .

An arbitrary point

P (x, y)

lies on the circle.

circle

Since the radius is

2,

the distance between

O (0, 0)

and

P (x, y)

is also

2 .

This information can be substituted into the Distance Formula.

d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

SubstituteValues

Substitute values

2 = (x - 0)^{2} + (y - 0)^{2}

Simplify

Subtract terms

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

$LHS^{2} = RHS^{2}$

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

Rearrange equation

x^{2} + y^{2} = 4

The equation of the given circle was obtained.

$x^{2} + y^{2} = 4$

Explore

Standard Equation of a Circle

The result previously obtained can be generalized to find the equation of a circle with a certain center and given radius.

By using this formula, the standard equation of any circle drawn on a coordinate plane can be written.

Example

Finding the Standard Equation of a Circle

Tearrik has one last problem to solve before going to a BBQ. He needs to find the standard equation of the circle shown below.

A circle on a coordinate plane

Tearrik remembers that the standard equation of a circle is $(x - h)^{2} + (y - k)^{2} = r^{2} .$ However, he does not remember how to find the values of $h,$ $k,$ and $r .$ Help Tearrik get to the BBQ by finding these values!

Hint

The center of the circle is $(h, k)$ and its radius $r .$

Solution

In the standard equation of a circle,

(h, k)

is the center of the circle. Therefore, by noting the coordinates of the center in the diagram, the values of

h

and

k

can be determined. Furthermore, the value of

r

is given by the distance between the center and any point on the circle.

The center of the circle can be seen to have the coordinates

(- 1, 2) .

Therefore, it can be stated that

h = - 1

and

k = 2 .

It can also be seen that the radius of the circle is

r = 5 .

By substituting these values into the standard equation of a circle, the equation of the given circle can be obtained.

(x - h)^{2} + (y - k)^{2} = r^{2}

SubstituteValues

Substitute values

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

Simplify

$a - (- b) = a + b$

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

Calculate power

(x + 1)^{2} + (y - 2)^{2} = 25

Example

Finding the Center and Radius of a Circle

Just like her classmate, Zain has one last problem to solve before getting to go to the BBQ. She has been asked to identify the center and the radius of the circle whose standard equation is given below.

(x - 4)^{2} + (y + 3)^{2} = 25

Zain has also been asked to graph the circle on a coordinate plane. Help Zain get to the BBQ!

Hint

The standard equation of a circle is $(x - h)^{2} + (y - k)^{2} = r^{2} .$ Here, the center of the circle is $(h, k)$ and its radius is $r .$

Solution

The standard equation of a circle is shown below.

(x - h)^{2} + (y - k)^{2} = r^{2}

Here, the center of the circle is

(h, k)

and its radius is

r .

Therefore, it is convenient to rewrite the given equation to match this format.

(x - 4)^{2} + (y + 3)^{2} = 25

Rewrite

3

as

- (- 3)

and Rewrite

25

as

5^{2}

$a + b = a - (- b)$

(x - 4)^{2} + (y - (- 3))^{2} = 25

Write as a power

(x - 4)^{2} + (y - (- 3))^{2} = 5^{2}

From the obtained equation, the center of the circle can be identified as

(4, - 3)

and its radius as

5 .

With this information, the circle can be drawn on a coordinate plane.

Closure

Determining a Point on the Circumference of a Circle

The challenge presented at the beginning of this lesson can be solved by writing the equation of the circle.

On a coordinate plane, a circle centered at the origin with radius $5$ was drawn. Also, a point on the circle with $x -$ coordinate $1$ was plotted.

circle and point

By writing the standard equation of the circle, find the

y -

coordinate of

P .

Write the answer as an exact value.

Hint

The standard equation of a circle is $(x - h)^{2} + (y - k)^{2} = r^{2},$ where $(h, k)$ is the center and $r$ the radius.

Solution

Recall the standard equation of a circle.

(x - h)^{2} + (y - k)^{2} = r^{2}

Here,

(h, k)

is the center and

r

the radius of the circle. Knowing that the center of the circle drawn in the diagram is

(0, 0)

and that its radius is

5,

its standard equation can be written.

(x - h)^{2} + (y - k)^{2} = r^{2}

SubstituteValues

Substitute values

(x - 0)^{2} + (y - 0)^{2} = 5^{2}

Simplify

Subtract terms

x^{2} + y^{2} = 5^{2}

Calculate power

x^{2} + y^{2} = 25

Recall that the

x -

coordinate of

P

is

1 .

Therefore, to find its

y -

coordinate, this value can be substituted into the equation of the circle.

x^{2} + y^{2} = 25

$x = 1$

1^{2} + y^{2} = 25

Solve for

y

$1^{a} = 1$

1 + y^{2} = 25

$LHS - 1 = RHS - 1$

y^{2} = 24

$LHS = RHS$

y = \pm 24

SplitIntoFactors

Split into factors

y = \pm 4 (6)

$a \cdot b = a \cdot b$

y = \pm 46

Calculate root

y = \pm 26

The

y -

coordinate of

P

can be either

26

or

- 26 .

However, from the diagram it can be observed that

P

is located in Quadrant I, where all values of the

y -

variable are positive. Therefore, the

y -

coordinate of

P

is

26 .

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