Topics
11. Circles
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United States of America
Grades 9-12

11.07 Circles in the coordinate plane

Lesson
Practice

Introduction

On the coordinate plane, we can find the equation of a circle using concepts we learned in  10.01 Distance and the coordinate plane  . When given in standard form, the equation highlights key features of the circle. If the equation is not in standard from, we will need to complete the square like we did in Algebra 1 lesson  11.03 Solving quadratic equations using square roots  .

Circles in the coordinate plane

All points on a circle are the same distance from the center. The radius tells us the distance from the center to any point on the circle.

Exploration

Consider the circle with a radius of 13 units shown below:

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  1. Verify the point \left(-11,4\right) lies on the circle.

  2. Write an equation that would allow you to find any point, \left(x,y\right), on the circle.

  3. Rewrite your equation to represent any circle with a center at \left(h,k\right) and radius, r.

The standard form of the equation of a circle is

\displaystyle \left(x-h\right)^2+\left(y-k\right)^2=r^2
\bm{r}
radius of the circle
\bm{\left(h,k\right)}
center of the circle
\bm{\left(x,y\right)}
coordinates of any point on the circle

To check whether a point \left(x_1,y_1\right) is inside, on or outside a circle, we can compare the distance between that point and the center of the circle to the value of the radius.

Using the Pythagorean theorem, we can write these conditions as:

  • If \left(x_1-h\right)^2+\left(y_1-k\right)^2<r^2 then \left( x_1,y_1 \right) is inside the circle
  • If \left(x_1-h\right)^2+\left(y_1-k\right)^2=r^2 then \left( x_1,y_1 \right) is on the circle
  • If \left(x_1-h\right)^2+\left(y_1-k\right)^2>r^2 then \left( x_1,y_1 \right) is outside the circle

Notice that these conditions are the same as substituting the point into the equation of the circle and comparing the values on each side.

An equation for a circle that has been expanded, referred to as expanded form of the equation of a circle, will be of the form x^2+y^2+ax+by+c=0

To convert this back into standard form, we can complete the square. Before completing the square, we need to rewrite the equation so the x-terms are together, the y-terms are together, and the constant is on the opposite side of the equation.x^2+ax+y^2+by=-c After it is in this form, we can complete the square for both the x-terms and the y-terms.

Examples

Example 1

Derive the equation of a circle with center \left(h,k\right) and radius r.

Worked Solution
Create a strategy

A circle is comprised of an infinite number of points that are equidistant from the center, \left(h,k\right). The distance from the center to any of these points is the length of the radius, r.

x
y

We can derive the equation of a circle by finding the length of r.

Apply the idea
x
y

We can draw a right triangle with a radius as the hypotenuse to see where the standard form for the equation of a circle comes from.

The length of the horizontal leg is \left |x-h\right |

The length of the vertical leg is \left | y-k\right |

By the Pythagorean theorem, {\left(x-h\right)^2+\left(y-k\right)^2=r^2}

Reflect and check

We could have also used the distance formula to find the length of r since the distance formula comes from the Pythagorean theorem.

\displaystyle r\displaystyle =\displaystyle \sqrt{\left(x-h\right)^2+\left(y-k\right)^2}
\displaystyle r^2\displaystyle =\displaystyle \left(x-h\right)^2+\left(y-k\right)^2

Example 2

Write the equation of the circle whose endpoints of a diameter are \left(-1.5, 4\right) and \left(4.5,-2\right).

Worked Solution
Create a strategy

To find the equation of the circle, we need to know the center of the circle and the length of the radius. The diameter of the circle passes through the center and is twice the radius. From this, we also know that the center is the midpoint of the diameter.

Apply the idea

Recall the midpoint formula: \left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right) Substituting the values from the endpoints of the diameter, we find the center of the circle to be:

\displaystyle \left(\dfrac{-1.5+4.5}{2},\dfrac{4+-2}{2}\right)\displaystyle =\displaystyle \left(1.5,1\right)

Next, we need to find the length of the radius which we can do by finding the distance from the center to either of the endpoints of the diameter. Using the first endpoint and the center, we find the radius to be:

\displaystyle r\displaystyle =\displaystyle \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}State the distance formula
\displaystyle =\displaystyle \sqrt{\left(-1.5-1.5\right)^2+\left(4-1\right)^2}Substitute values from the points
\displaystyle =\displaystyle \sqrt{\left(-3\right)^2+\left(3\right)^2}Evaluate the subtraction
\displaystyle =\displaystyle \sqrt{9+9}Evaluate the exponents
\displaystyle =\displaystyle \sqrt{18}Evaluate the addition

Now, we can substitute the center and radius into the standard form of a circle.

The equation of the circle will be:\left(x-1.5\right)^2+\left(y-1\right)^2=\left(\sqrt{18}\right)^2

which can be simplified to: \left(x-1.5\right)^2+\left(y-1\right)^2=18

Reflect and check

Using technology to graph the circle, we can see that the equation of the circle is correct because the given endpoints do lie on a diameter of the circle.

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Example 3

A circle has the equation x^2+y^2-6x+4y-12=0.

a

Rewrite the equation of the circle in standard form by completing the square.

Worked Solution
Create a strategy

To rewrite the equation in standard form, we need to complete the square twice, once for the x-terms and again for the y-terms. Before completing the square, we should group the x- and y-terms together and move the constant to the other side.

x^2-6x+y^2+4y=12

Apply the idea
\displaystyle x^2-6x+y^2+4y\displaystyle =\displaystyle 12Given equation
\displaystyle x^2-6x+9 +y^2+4y\displaystyle =\displaystyle 12+9Complete the square for the x-terms
\displaystyle \left(x-3\right)^2 +y^2+4y\displaystyle =\displaystyle 21Factor the perfect square trinomial
\displaystyle \left(x-3\right)^2 +y^2+4y+4\displaystyle =\displaystyle 21+4Complete the square for the y-terms
\displaystyle \left(x-3\right)^2 +\left(y+2\right)^2\displaystyle =\displaystyle 25Factor the perfect square trinomial
Reflect and check

Now that the equation is in standard form, we can identify the center and radius. Notice the equation can be written as \left(x-3\right)^2 +\left(y+2\right)^2=5^2 to easily identify the radius.

The center of the circle is at \left(3,-2\right), and the radius is 5 units in length.

b

Determine whether the point \left(1,1\right) is inside, outside, or on the circle.

Worked Solution
Create a strategy

We want to find the distance between the point \left( 1,1 \right) and the center of the circle and compare it to the radius. We can do this by using the standard form found in part (a) to compare the lengths.

Apply the idea

We will substitute the x- and y-value of the given point into the left side of the equation we found in part (a).

\displaystyle \left(x-3\right)^2+\left(y+2\right)^2\displaystyle =\displaystyle \left(1-3\right)^2+\left(1+2\right)^2Substitute x=1 and y=1
\displaystyle =\displaystyle \left(-2\right)^2+\left(3\right)^2Evaluate the parentheses
\displaystyle =\displaystyle 4+9Evaluate the exponents
\displaystyle =\displaystyle 13Evaluate the addition
\displaystyle 25\displaystyle >\displaystyle 13Compare to r^2

Since the distance between the point \left(1,1\right) is less than the radius, the point is inside the circle.

Reflect and check

We can use technology to graph the circle and check that the point does lie within the circle.

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Example 4

Darnell shines a torch at a wall which lights up a circular region with a diameter of 4 meters. The center of the light is positioned 3 meters above the ground, and 5 meters horizontally from the left side of the wall.

a

Let the bottom left corner of the wall be the origin. Determine the equation of the circle which describes the edge of lighted area.

Worked Solution
Create a strategy

We are told that the center of the light is 5 meters from the left side of the wall, which is 5 meters to the right of the origin. It is also 3 meters above the ground, which is 3 meters up from the origin.

We are also given that the diameter of the lighted area is 4 meters. We can use this information to determine the equation of the circle of the lighted area.

Apply the idea

The center of the lighted area will have the coordinates \left(5,3\right), and the radius will be 2 meters.

The equation of the circle which describes the edge of the lighted area is\left(x-5\right)^2+\left(y-3\right)^2=2^2 which we can simplify to \left(x-5\right)^2+\left(y-3\right)^2=4

b

Yvonne has a height of 1.66 meters and is standing against the wall, 5 meters from the left side. Determine if any part of Yvonne is in the lighted area.

Worked Solution
Create a strategy

Notice that the x-coordinate of Yvonne's position is the same of the x-coordinate of the center of the circle. This means we only need to determine if the bottom edge of the circle will reach Yvonne.

Apply the idea

Since the radius of the lighted area is 2 meters, the lowest point that the light reveals will be 2 meters below the y-value of the center. The lighted area will begin from a height of 3-2=1 meter.

Since both Yvonne and the center of the light are positioned 5 meters from the left side of the wall, and Yvonne is taller than 1 meter, part of Yvonne will be in the lighted area.

c

Kayoko is standing against the wall, 6 meters from the left side of the wall. Determine the greatest height that Kayoko can be without being in the lighted area. Round your answer to the nearest centimeter.

Worked Solution
Create a strategy

For Kayoko to not be in the lighted area, her height must be less than or equal to the lowest point that the light reveals at 6 meters from the left side of the wall. In other words, the greatest height Kayoko can have without being seen is equal to the smallest y-value of the edge of the lighted area when x=6.

Apply the idea

We will substitute x=6 into the standard form of the equation we found in part (a) and solve for the height, y.

\displaystyle \left(x-5\right)^2+\left(y-3\right)^2\displaystyle =\displaystyle 4Standard form of the equation of the circle
\displaystyle \left(6-5\right)^2+\left(y-3\right)^2\displaystyle =\displaystyle 4Substitute x=6
\displaystyle \left(y-3\right)^2\displaystyle =\displaystyle 4-\left(6-5\right)^2Subtract \left(6-5\right)^2 from both sides
\displaystyle \left(y-3\right)^2\displaystyle =\displaystyle 3Evaluate the right-hand side of the equation
\displaystyle y-3\displaystyle =\displaystyle \pm\sqrt{3}Square root property
\displaystyle y\displaystyle =\displaystyle \pm\sqrt{3}+3Add 3 to both sides
\displaystyle y=4.73,\, y\displaystyle =\displaystyle 1.26Evaluate the expression, rounding down to the nearest centimeter

Taking the smallest y-value, we get that Kayoko can be up to 1.26 meters or 126 centimeters tall without being in the lighted area.

Reflect and check

Recall there are 100 centimeters in 1 meter which is why we rounded to 2 decimal places.

In meters, the numbers may seem small for heights. But if we converted 1.26 meters to feet, it would be about 4 feet and 2 inches which is a reasonable height for a student.

Idea summary

The standard form of the equation of a circle is

\displaystyle \left(x-h\right)^2+\left(y-k\right)^2=r^2
\bm{r}
radius of the circle
\bm{\left(h,k\right)}
center of the circle
\bm{\left(x,y\right)}
coordinates of any point on the circle

To convert an equation in expanded form to standard form, we must complete the square for the x-terms and the y-terms.

Outcomes

G.GPE.A.1

Derive the equation of a circle of given center and radius using the Pythagorean theorem; complete the square to find the center and radius of a circle given by an equation.

G.GPE.B.4

Use coordinates to prove simple geometric theorems algebraically.

G.MG.A.1

Use geometric shapes, their measures, and their properties to describe objects.

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