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Geo

Geometry View details

8. Equation of a Circle

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Chapter 5

8.

Equation of a Circle

Understanding the coordinates of a circle is vital in geometry. The position and size of a circle in a plane can be described using its coordinates. The standard equation of a circle is a mathematical tool that helps in representing a circle accurately on a coordinate plane. By mastering this equation, one can determine essential aspects of a circle such as its center and radius. In real-world applications, from architecture to satellite trajectory, knowing the coordinates and the standard equation enables precise circular designs and predictions.

Lesson Settings & Tools

	9 Theory slides
	11 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions

Equation of a Circle

Slide of 9

Circles have been previously studied in this course. In this lesson, the equation of a circle on a coordinate plane will be derived. Furthermore, when given the equation of a circle, its center and radius will be found, and its graph will be drawn.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Circles
Distance Formula
Completing the square
Solving equations

Try your knowledge on these topics.

a By using the Distance Formula, find the distance between points

P

and

Q .

b Identify the center and find the radius of the circle below.

c By completing the square, determine the values of

h

and

k .

x^{2} + 6 x + 11 = (x - h)^{2} + k

d Solve the linear equation.

3 x + 8 = 53

e Solve the quadratic equation.

x^{2} + 5 x = 50

Write the smaller solution first.

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.

On a coordinate plane, consider a circle with radius $r$ and center $(h, k) .$

standard equation of a circle

The standard equation of the above circle is given below.

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

Proof

On the diagram below, consider an arbitrary point on a circle with coordinates

(x, y) .

As it can be seen in the diagram, the distance between the center

(h, k)

and the point

(x, y)

is

r .

This information can be substituted into the Distance Formula.

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

SubstituteValues

Substitute values

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

▼

Simplify

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

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

Rearrange equation

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

The equation of a circle with radius

r

and center

(h, k)

was obtained by using the Distance Formula.

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.

Example

Rewriting the Equation of a Circle

Sometimes the equation of a circle needs a significant change to be rewritten as the standard equation of a circle. Typically in those cases, the equation can be rewritten by completing the square.

To be allowed to help design her schools basketball court, Dominika was asked to identify the center and the radius of the circle whose equation is given below.

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

By rewriting the above equation as the standard equation of a circle, identify the center and the radius. If any answer is an irrational number, write its exact value.

Hint

Add a number to the expression $x^{2} - 2 x$ so that it becomes a perfect square trinomial.

Solution

To rewrite the given equation as the standard equation of a circle, a method called completing the square can be used. This method consists in writing an expression as a perfect square trinomial so that it can be factored as the square of a binomial. First, the expression

x^{2} - 2 x

will be considered.

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

Ignoring the negative sign, the coefficient of the variable

x

is

2 .

This coefficient is commonly called

b .

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

Next, the term

(\frac{b}{2})^{2}

will be added to and subtracted from the expression. Note that adding and subtracting the same value does not modify the expression. Since

b = 2,

the value of

(\frac{2}{2})^{2}

can be calculated.

(\frac{2}{2})^{2}

▼

Simplify

$\frac{a}{a} = 1$

1^{2}

$1^{a} = 1$

1

Therefore,

1

will be added to and subtracted from

x^{2} - 2 x .

Then, the resulting perfect square trinomial will be factored and written as the square of a binomial.

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

Identity Property of Addition

x^{2} - 2 x + 0 + y^{2} = 4

Rewrite $0$ as $1 - 1$

x^{2} - 2 x + 1 - 1 + y^{2} = 4

▼

a^{2} - 2 a b + b^{2} = (a - b)^{2}

Identity Property of Multiplication

x^{2} - 2 x (1) + 1 - 1 + y^{2} = 4

Write as a power

x^{2} - 2 x (1) + 1^{2} - 1 + y^{2} = 4

FacNegPerfectSquare

$a^{2} - 2 a b + b^{2} = (a - b)^{2}$

(x - 1)^{2} - 1 + y^{2} = 4

The process of completing the square is now finished. Finally, to obtain the standard equation of the circle, the number

1

will be added to both sides of the equation and

y

will be written as

y - 0 .

Also, the resulting number on the right-hand side will be expressed as a square.

(x - 1)^{2} - 1 + y^{2} = 4

▼

Simplify

$LHS + 1 = RHS + 1$

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

Rewrite $y$ as $y - 0$

(x - 1)^{2} + (y - 0)^{2} = 5

Rewrite $5$ as $(5)^{2}$

(x - 1)^{2} + (y - 0)^{2} = (5)^{2}

The standard equation of the circle was obtained. The center can be identified as

(1, 0)

and the radius as

5 .

Explore

Rewriting the Equation of One More Circle

This time, Dominika wants to play basketball on Sunday. Her father will be okay with that only if she completes her math homework. To do so, Dominika has to identify the center and the radius of the circle whose equation is given below.

x^{2} + 6 x + y^{2} - 4 y = - 3

By rewriting the above equation as the standard equation of a circle, identify the center and the radius. If any of the answers is an irrational number, write its exact value.

Hint

Complete the square for the $x -$ and the $y -$ variable.

Solution

To rewrite the given equation as the standard equation of a circle, the process of completing the square needs to be performed for both variables.

x^{2} + 6 x + y^{2} - 4 y = - 3

Identity Property of Addition

x^{2} + 6 x + 0 + y^{2} - 4 y + 0 = - 3

Rewrite $0$ as $9 - 9 & 4 - 4$

x^{2} + 6 x + 9 - 9 + y^{2} - 4 y + 4 - 4 = - 3

▼

a^{2} \pm 2 a b + b^{2} = (a \pm b)^{2}

SplitIntoFactors

Split into factors

x^{2} + 2 (3) x + 9 - 9 + y^{2} - 2 (2) y + 4 - 4 = - 3

CommutativePropMult

Commutative Property of Multiplication

x^{2} + 2 x (3) + 9 - 9 + y^{2} - 2 y (2) + 4 - 4 = - 3

Write as a power

x^{2} + 2 x (3) + 3^{2} - 9 + y^{2} - 2 y (2) + 2^{2} - 4 = - 3

$a^{2} \pm 2 a b + b^{2} = (a \pm b)^{2}$

(x + 3)^{2} - 9 + (y - 2)^{2} - 4 = - 3

▼

Simplify

Subtract terms

(x + 3)^{2} + (y - 2)^{2} - 13 = - 3

$LHS + 13 = RHS + 13$

(x + 3)^{2} + (y - 2)^{2} = 10

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

(x - (- 3))^{2} + (y - 2)^{2} = 10

Rewrite $10$ as $(10)^{2}$

(x - (- 3))^{2} + (y - 2)^{2} = (10)^{2}

The standard equation of the circle was obtained. The center can be identified as

(- 3, 2)

and the radius as

10 .

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 .

Equation of a Circle

Exercise 3.1

Find the equation of the circumscribed circle to the triangle with vertices in

(- 8, 1),

(- 8, 6),

and

(4, 1) .

To write the equation of the circumscribed circle we need to find the circles center, which is the same thing as the triangle's circumcenter, as well as the circle's radius. Then we will substitute these values into the standard equation of a circle.

Finding the Circumcenter

Let's begin by drawing the triangle.

The circumscribed circle to this triangle will have its center in the circumcenter. To find that we need to construct a perpendicular bisector to at least two of the triangle's sides. Notice that this is a right triangle which means the circumcenter will be on the hypotenuse.

The circumcenter is the point of intersection of the perpendicular bisectors. This point is (- 2,3.5).

Finding the Radius

The circumscribed circle will contain each of the triangle's vertices. To find the radius of the circle we need to find the distance from its center to one of the vertices.

Let's find the distance r using the Distance Formula.

Writing the Equation

Let's recall the standard equation of a circle. (x- h)^2+(y- k)^2= r^2 In this form, ( h, k) is the center of the circle and r is its radius. We have found that the center of the circle is ( - 2, 3.5) and that its radius is 6.5. Let's use this to write the equation. (x-( - 2))^2+(y- 3.5)^2= 6.5^2 ⇕ (x+2)^2+(y-3.5)^2=42.25

Equation of a Circle

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