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.
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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.
3x+8=53
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
Try your knowledge on these topics.
a By using the Distance Formula, find the distance between points P and Q.
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b Identify the center and find the radius of the circle below.
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c By completing the square, determine the values of h and k.
x^2+6x+11=(x-h)^2+k
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d Solve the linear equation.
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e Solve the quadratic equation.
x^2+5x=50 Write the smaller solution first.
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Investigating Points on a Circle's Circumference
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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. 
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.
The equation of the given circle was obtained.
d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)
SubstituteValues
Substitute values
2=sqrt(( x- 0)^2+( y- 0)^2)
▼
Simplify
x^2+y^2=4
x^2+y^2=4
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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).
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.
The equation of a circle with radius r and center (h,k) was obtained by using the Distance Formula.
d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)
SubstituteValues
Substitute values
r=sqrt(( x- h)^2+( y- k)^2)
▼
Simplify
(x-h)^2+(y-k)^2=r^2
Example
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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.
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!
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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.
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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!
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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.
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.
(x-4)^2+(y+3)^2=25
▼
Rewrite 3 as -(- 3) and Rewrite 25 as 5^2
(x- 4)^2+(y-( - 3))^2= 5^2
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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^2-2x+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.
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Hint
Add a number to the expression x^2-2x 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^2-2x will be considered.
x^2-2x+y^2=4
Ignoring the negative sign, the coefficient of the variable x is 2. This coefficient is commonly called b.
x^2- 2x+y^2=4
Next, the term ( b2)^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 ( 22)^2 can be calculated.
Therefore, 1 will be added to and subtracted from x^2-2x. Then, the resulting perfect square trinomial will be factored and written as the square of a binomial.
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.
The standard equation of the circle was obtained. The center can be identified as ( 1, 0) and the radius as sqrt(5).
x^2-2x+y^2=4
IdPropAdd
Identity Property of Addition
x^2-2x+0+y^2=4
Rewrite
Rewrite 0 as 1-1
x^2-2x+ 1- 1+y^2=4
▼
a^2-2ab+b^2=(a-b)^2
IdPropMult
Identity Property of Multiplication
x^2-2x(1)+1-1+y^2=4
WritePow
Write as a power
x^2-2x(1)+1^2-1+y^2=4
FacNegPerfectSquare
a^2-2ab+b^2=(a-b)^2
(x-1)^2-1+y^2=4
(x-1)^2-1+y^2=4
(x- 1)^2+(y- 0)^2=( sqrt(5))^2
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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+6x+y^2-4y=- 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.
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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.
The standard equation of the circle was obtained. The center can be identified as ( - 3, 2) and the radius as sqrt(10).
x^2+6x+y^2-4y=- 3
IdPropAdd
Identity Property of Addition
x^2+6x+0+y^2-4y+0=- 3
Rewrite
Rewrite 0 as 9-9 & 4-4
x^2+6x+ 9- 9+y^2-4y+ 4- 4=- 3
▼
a^2± 2ab+b^2=(a± 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+2x(3)+9-9+y^2-2y(2)+4-4=- 3
WritePow
Write as a power
x^2+2x(3)+3^2-9+y^2-2y(2)+2^2-4=- 3
a^2± 2ab+b^2=(a± b)^2
(x+3)^2-9+(y-2)^2-4=- 3
(x-( - 3))^2+(y- 2)^2=( sqrt(10))^2
Closure
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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.
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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.
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.
The y-coordinate of P can be either 2sqrt(6) or - 2sqrt(6). 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 2sqrt(6). 
Equation of a Circle
Exercise 2.1
Consider the following circle. x^2+y^2-4x=5
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Which of the following diagrams represents its graph?
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Let's start by recalling the standard equation of a circle. (x- h)^2+(y- k)^2= r^2 We will rewrite the given equation to match this form. In this case, we will need to complete the square.
We can now simplify this a little. (x- 2)^2+(y- 0)^2= 3^2 ⇓ (x-2)^2+y^2=9
Let's have a look at the standard equation of a circle.
(x- h)^2+(y- k)^2=r^2
Here, the center is the point ( h, k). We can use this to identify the center of our circle.
(x- 2)^2+(y- 0)^2=3^2
The circle has its center in ( 2, 0).
The radius is written as r in the standard equation of a circle.
(x-h)^2+(y-k)^2= r^2
Let's compare this with the expression for our circle.
(x-2)^2+(y-0)^2= 3^2
We can see that our circle has a radius of 3.
In Part B, we found that the circle has its center in ( 2, 0). In Part C, we learned that the circle has a radius of 3. By using this information we can draw its graph.
We can now conclude that graph iv represents the circle.
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Consider the following circle. x^2+y^2+4x+12y=- 15
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Which of the following diagrams represents its graph?
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Let's recall the standard equation of a circle. (x- h)^2+(y- k)^2= r^2 We will rewrite the given equation to match this form. In this case, we will need to complete the square twice — once for each variable.
We can now simplify this a little. (x-( - 2))^2+(y-( - 6))^2= 5^2 ⇓ (x+2)^2+(y+6)^2=25
Let's once more look into the standard equation of a circle.
(x- h)^2+(y- k)^2=r^2
Here, the center is the point ( h, k). We can use this to identify the center of our circle.
(x-( - 2))^2+(y-( - 6))^2=5^2
The circle has its center in ( - 2, - 6).
The radius is denoted r in the standard equation of a circle.
(x-h)^2+(y-k)^2= r^2
Let's compare this with the expression for our circle.
(x-(- 2))^2+(y-(- 6))^2= 5^2
We can see that our circle has a radius of 5.
In Part B we found that the circle has its center in ( - 2, - 6). In Part C we learned that the circle has a radius of 5. By using this information we can draw its graph.
We can now identify that graph i represents the circle.
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A circle has its center in (-2,3) and an x-intercept at (-5,0). Find its other x-intercept.
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We will first find the circle's radius. When we know the center and the radius we can write the circle's equation. Finally, we will use the equation to find the circle's second x-intercept.
Finding the Radius
Let's recall the Distance Formula. d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2) We find the radius of our circle by substituting the center and the x-intercept into this equation, then evaluating the right-hand side.
The circle has a radius of sqrt(18) units.
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. The exercise tells us that the center of the circle is ( - 2, 3) and we have calculated its radius and found that it is sqrt(18) units. Let's use this knowledge to write the equation. (x-( - 2))^2+(y- 3)^2=( sqrt(18))^2 ⇕ (x+2)^2+(y-3)^2=18
Finding the x-intercept
We find the circle's x-intercepts by substituting 0 for y into the equation and solving for x.
The circle has two x-intercepts. One of them (- 5,0) was given to us in the exercise. The other is at (1,0).
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A circle has its center in (3,1) and a radius of 4. Find its y-intercept(s). Answer in exact form.
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We will first write the equation of the circle. Then we will use this equation to find the circle's y-intercept(s).
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. The center of the circle is ( 3, 1) and it has a radius of 4. Let's use this to write the equation. (x- 3)^2+(y- 1)^2= 4^2 ⇕ (x-3)^2+(y-1)^2=16
Finding the y-intercepts
We find the circle's y-intercept(s) by substituting 0 for x into the equation, then solve for y.
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Equation of a Circle
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