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12. Circles
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Geometry

12.04 Equations of circles

Lesson
Practice

Equations of circles

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.

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, \left(x,y\right), is the length of the radius, r.

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

Apply the idea
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y

We can begin by drawing a right triangle with a radius as the hypotenuse. Then, we need to find the lengths of each leg of the right triangle.

The lengths of the lengths are simply the horizontal and vertical distances between the center, \left(h,k\right), and a point on the circle, \left(x,y\right).

A line from point (h,k) to point (x,y) is drawn on the coordinate plane. The horizontal distance from the y-axis to point (h,k) is labeled h. The horizontal distance from the y-axis to point (x,y) is labeled x.

The horizontal distance between the two points is the absolute value of the difference between the x-values.

|x-h|

A line from point (h,k) to point (x,y) is drawn on the coordinate plane. The vertical distance from the x-axis to point (h,k) is labeled k. The vertical distance from the x-axis to point (x,y) is labeled y.

The vertical distance between the two points is the absolute value of the difference between the y-values.

|y-k|

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y

Now that the lengths of the legs of the triangle are known, we can use the Pythagorean theorem to find the distance between the two points.

\left|x-h\right|^{2} + \left|y-k\right|^{2} = r^{2}

Since squaring a expression will always result in a positive value, the absolute value bars are not necessary. Therefore, the equation of a circle with center \left(h,k\right) and radius r is \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

Consider the circle shown.

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a

State the coordinates of the center.

Worked Solution
Create a strategy

Notice that this circle has several points with integer coordinates. By connecting the points that lie directly across the circle, we can assume these lines will be close to the diameter. Since we know a diameter passes through the center, the point where these lines intersect will be the center.

Apply the idea
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The center of the circle is at the point \left(2, 0 \right).

b

State the radius of the circle.

Worked Solution
Create a strategy

The length of a circle's radius is the distance from the center point to any point on the circumference of the circle.

Apply the idea

In the previous part, we identified the center of the circle, \left(2, 0\right), and several other points on the circle. To find the radius, we can count the number of units the center is from one of the points on the circle.

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The distance from \left(2,5\right) to \left(2,0\right) is \left|5-0\right|=5, so the length of the radius is 5 units.

c

State the diameter of the circle.

Worked Solution
Create a strategy

The diameter of a circle is any straight line segment that passes through the center of the circle and whose endpoints lie on the circrumference of the circle. The diameter is defined as twice the length of the radius of the circle.

Apply the idea

We found in the previous part that the radius of the circle is 5 units. Multiplying the radius by 2 will give us the diameter. This circle has a diamemter length of 10 units.

d

State the equation of the circle.

Worked Solution
Create a strategy

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

Apply the idea

The center coordinates are \left(h,k\right)=\left(0,2\right) and the length of the radius it r=5.

Substituting these values into the standard form of the equation of a circle, we get \left(x-2 \right)^{2} + \left(y-0\right)^{2} = 5^{2}

Simplifying the equation, we get \left(x - 2 \right)^{2} +y^{2} = 25

Reflect and check

We can use technology to graph the circle represented by our equation and confirm that it matches the original graph.

Example 3

Write the equation of the circle with the given conditions.

a

Center at \left(-1, 2 \right) and a radius of 4

Worked Solution
Create a strategy

To write the equation of a circle, we use the standard form \left(x-h\right)^{2}+ \left(y-k\right)^{2}=r^{2}, where \left(h,k\right) represents the center of the circle and r is the radius. Given the center \left(-1,2\right) and radius 4, we can directly substitute these values into the equation.

Apply the idea

Substituting the values of h = -1, k = 2 and r = 4 into standard form of the circle equation gives us:

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

Simplifying further, we get:

\left( x + 1 \right)^{2} + \left(y - 2\right)^{2} = 16

Reflect and check

Recall the standard form of the equation of a circle is \left(x - h \right)^{2} + \left(y-k \right)^{2}= r^{2}. The values of h and k can be directly related to translations of circles from the parent circle, x^{2}+y^{2}=r^{2}, centered at the origin.

The value of h represents a horizontal translation. In this case, h shifts the circle left 1 unit. The value of k represents a vertical translation. In this case, k shifts the circle up 2 units. Combining the movements, our circle was translated left 1 unit and up 2 units from the origin.

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b

Center at \left(2, 4 \right) and a point on the circle at \left(-2, 1 \right)

Worked Solution
Create a strategy

To find the equation of a circle, we need to use the formula \left(x-h\right)^{2}+ \left(y-k\right)^{2}=r^{2}, where \left(h,k \right) is the center of the circle and r is the radius. The radius can be calculated using the distance formula between two points, which in this case are the center and the given point on the circle.

Apply the idea

First, we calculate the radius r by finding the distance between the center, \left(h, k \right) = \left(2, 4 \right), and the given point on the circle, \left(x_1, y_1 \right) = \left(-2,1 \right).

\displaystyle r\displaystyle =\displaystyle \sqrt{\left(x_2-x_1\right)^{2}+\left(y_2-y_1\right)^{2}}Distance formula
\displaystyle =\displaystyle \sqrt{\left(-2-2\right)^{2}+\left(1-4\right)^{2}}Subsitute the values of the points
\displaystyle =\displaystyle \sqrt{\left(-4\right)^{2}+\left(-3\right)^{2}}Evaluate the subtraction
\displaystyle =\displaystyle \sqrt{16+9}Evaluate the squares
\displaystyle =\displaystyle \sqrt{25}Evaluate the addition
\displaystyle =\displaystyle 5Square root of a perfect square

With r=5, we can now write the equation of the circle:

\left(x-2\right)^{2} + \left(y - 4 \right)^{2} = 5^{2}

\left(x - 2\right)^{2} + \left(y - 4 \right)^{2} = 25

Reflect and check

We have the equation of the circle to be \left(x-2\right)^{2} + \left(y-4 \right)^{2}=25. To check the accuracy of the equation, we can substitute the coordinates of the given point \left(-2,1 \right) to make sure they satisfy this equation:

\displaystyle (x-2)^{2}+(y-4)^{2}\displaystyle =\displaystyle 25Equation of the circle
\displaystyle (-2-2)^{2}+(1-4)\displaystyle =\displaystyle 25Substitute \left(x,y\right)=\left(-2,1\right)
\displaystyle 16 + 9\displaystyle =\displaystyle 25Simplify
\displaystyle 25\displaystyle =\displaystyle 25\checkmark

This point does satisfy the equation, showing the equation we found for the circle is correct.

c

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 coordinates of 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 the equation 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 4

Darnell shines a flashlight 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

Outcomes

G.PC.4

The student will solve problems in the coordinate plane involving equations of circles.

G.PC.4bi

Solve problems in the coordinate plane involving equations of circles: i) given a graph or the equation of a circle in standard form, identify the coordinates of the center of the circle;

G.PC.4bii

Solve problems in the coordinate plane involving equations of circles: ii) given the coordinates of the endpoints of a diameter of a circle, determine the coordinates of the center of the circle.

G.PC.4biii

Solve problems in the coordinate plane involving equations of circles: iii) given a graph or the equation of a circle in standard form, identify the length of the radius or diameter of the circle.

G.PC.4bv

Solve problems in the coordinate plane involving equations of circles: v) given the coordinates of the center and the coordinates of a point on the circle, determine the length of the radius or diameter of the circle; and

G.PC.4bvi

Solve problems in the coordinate plane involving equations of circles: vi) given the coordinates of the center and length of the radius of a circle, identify the coordinates of a point(s) on the circle.

G.PC.4ci

Determine the equation of a circle given: i) a graph of a circle with a center with coordinates that are integers;

G.PC.4cii

Determine the equation of a circle given: ii) coordinates of the center and a point on the circle;

G.PC.4ciii

Determine the equation of a circle given: iii) coordinates of the center and the length of the radius or diameter;

G.PC.4civ

Determine the equation of a circle given: iv) coordinates of the endpoints of a diameter.

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