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In order to access this I need to be confident with:

Coordinate PlaneSolving equations

Pythagorean theoremDistance formula

SubstitutionSquare roots

Here you will learn about the equation of a circle, including how to recognize the equation of a circle, form an equation of a circle given its radius and center, use the equation of a circle to find its center and radius, and solve problems involving the equation of a circle.

Students will first learn about the equation of a circle as part of geometry in high school.

The **equation of a circle** is x^2+y^2=r^2, where r represents the radius, with a center at 0,0 and (x-h)^2+(y-k)^2=r^2, where r represents the radius, with a center at (h,k).

The definition of a circle is a set of all points on a plane that are a fixed distance from a center. That distance is called the radius.

To understand the equation of a circle, consider the drawing below of a circle on a set of axes with the center at the origin.

Now consider a right angled triangle created when the radius of the circle is the hypotenuse of the right triangle.

- The horizontal line is the distance to the x coordinate
- The vertical line is the distance to the y coordinate
- The hypotenuse is the distance of the radius

You can now apply Pythagoras’ theorem to the above. Therefore, the general form of the equation of a circle centered at (0,0) is:

x^2+y^2=r^2

x = x coordinate

y= y coordinate

r = radius

For example,

Draw circle with equation x^2+y^2=9.

The circle has a center at (0,0).

9 represents r^2, so the radius r is 3.

But what about a circle that is not centered on the origin? Let’s look at the same circle as above, but in a different position.

The radius of the circle below is still 3, but now the center is at (1,1). This means all the points on the circle have shifted up 1 and right 1.

Since x^2+y^2=r^2 is true for all circles a (0,0), you can shift the circle back down to the origin.

To do that, you need to **shift all the ** \textbf{x} ** coordinates left ** \bf{1} on the x- axis and **shift all the ** \textbf{y} ** coordinates down ** \bf{1} on the y- axis to return this circle to the origin.

This leaves us with the equation (x-1)^2+(y-1)^2=9.

What about if this same circle was in another position?

The radius of the circle below is still 3, but now the center is at (-1,-1). This means all the points on the circle have shifted left 1 and down 1.

Since x^2+y^2=r^2 is true for all circles a (0,0), you need to **shift all the ** \textbf{x} ** coordinates right ** \bf{1} and **shift all the ** \textbf{y} ** coordinates up ** \bf{1} to return this circle to the origin. This leaves us with the equation (x+1)^2+(y+1)^2=9.

Therefore, the general form of the equation of any circle is:

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

h= x coordinate of center of the circle

k = y coordinate of center of the circle

r = radius of the circle

For example,

Draw the circle with equation (x-2)^2+(y-3)^2=4^2.

h = x coordinate of center = 2

k = y coordinate of center = 3

Coordinates of the center = (2,3)

r = length of radius = 4

For example,

What is the equation of the circle with a center of (2,5) and a radius of 10~{cm}?

The standard equation of a circle is (x-h)^2+(y-k)^2=r^2.

- The x -coordinate of center of the circle is 2
- The y -coordinate of the center of the circle is 5
- The radius is 10

So,

\begin{aligned} &(x-h)^{2}+(y-k)^{2}=r^{2} \\\\ &(x-2)^{2}+(y-5)^{2}=10^{2} \\\\ &(x-2)^{2}+(y-5)^{2}=100 \end{aligned}

How does this relate to high school math?

**Geometry – Expressing Geometric Properties with Equations (HSG.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.

Use this quiz to check your grade 5 to 8 students’ understanding of type of graphs. 15+ questions with answers covering a range of 5th to 8th grade type of graphs topics to identify areas of strength and support!

DOWNLOAD FREEUse this quiz to check your grade 5 to 8 students’ understanding of type of graphs. 15+ questions with answers covering a range of 5th to 8th grade type of graphs topics to identify areas of strength and support!

DOWNLOAD FREEIn order to use the equation of a circle:

**Write the general equation of a circle.****State any variables you know.****Substitute any values you know into the equation.****Use the information you have to solve the problem.****Clearly state the answer.**

What is the equation of the circle with a center at the origin?

**Write the general equation of a circle.**

x^2+y^2=r^2 for circles at the origin.

2**State any variables you know.**

The distance from the center (0,0) to the circle is 4.

Radius (r) = 4

3**Substitute any values you know into the equation.**

x^2+y^2=4^2

4**Use the information you have to solve the problem.**

Simplify the equation by squaring the radius.

\begin{aligned} &x^{2}+y^{2}=4^{2} \\\\ &x^{2}+y^{2}=16 \end{aligned}

5**Clearly state the answer.**

The equation of the circle is: x^2+y^2=16

What is the radius of the circle with the equation x^2+y^2=4?

**Write the general equation of a circle.**

x^2+y^2=r^2 for circles at the origin.

**State any variables you know.**

Radius, r, is unknown.

Equation of the circle is given as x^2+y^2=4.

**Substitute any values you know into the equation.**

You do not know any variables so we are unable to substitute here.

**Use the information you have to solve the problem.**

You know that the radius squared is equal to 4, so

\begin{aligned}
r^{2}&=4 \\\\
r&=\sqrt{4} \\\\
r&=2
\end{aligned}

Note: You only use the positive value as the radius’ measure of distance.

**Clearly state the answer.**

The radius is 2.

What is the equation of a circle with a radius of \sqrt5 and a center at the origin?

**Write the general equation of a circle.**

x^2+y^2=r^2 for circles at the origin.

**State any variables you know.**

Radius (r) = \sqrt5

**Substitute any values you know into the equation.**

x^{2}+y^{2}=(\sqrt{5})^{2}

**Use the information you have to solve the problem.**

Simplify the equation by squaring the radius.

\begin{aligned}
&x^{2}+y^{2}=(\sqrt{5})^{2} \\\\
&x^{2}+y^{2}=5
\end{aligned}

**Clearly state the answer.**

The equation of the circle is: x^2+y^2=5

What is the equation of the circle with diameter endpoints at (-5,3) and (9,3)?

**Write the general equation of a circle.**

(x-h)^2+(y-k)^2=r^2, where r represents the radius, with a center at (h,k).

**State any variables you know.**

The distance from (-5,3) to the point (9,3) on the circle is 14. This is the diameter of the circle. The length of the radius is half the length of the diameter.

Radius (r) = 7

The center is at the midpoint of the diameter. Counting over 7 from (-5,3) shows that the center is at (2,3).

\begin{aligned} & h=2 \\\\ & k=3 \end{aligned}

**Substitute any values you know into the equation.**

(x-2)^2+(y-3)^2=7^2

**Use the information you have to solve the problem.**

Simplify the equation by squaring the radius.

\begin{aligned} & (x-2)^2+(y-3)^2=7^2 \\\\ & (x-2)^2+(y-3)^2=49 \end{aligned}

**Clearly state the answer.**

The equation of the circle is: (x-2)^2+(y-3)^2=49

What is the center and radius of the circle with the equation

(x+2)^2+(y-5)^2=8?

**Write the general equation of a circle.**

(x-h)^2+(y-k)^2=r^2, where r represents the radius, with a center at (h,k).

**State any variables you know.**

The radius squared (r^2) = 8

h = -2

k = 5

**Substitute any values you know into the equation.**

You do not know any variables so we are unable to substitute here.

**Use the information you have to solve the problem.**

You know that the radius squared is equal to 8, so

\begin{aligned} & r^2=8 \\\\ & r=\sqrt{8} \\\\ & r=2 \sqrt{2} \end{aligned}

Note: You only use the positive value as the radius’ measure of distance.

**Clearly state the answer.**

The radius is 2\sqrt{2} and the center is at (-2,5).

What is the equation of a circle with a radius of \sqrt20 and a center at (4, -8)?

**Write the general equation of a circle.**

(x-h)^2+(y-k)^2=r^2 for circles at (h, k).

**State any variables you know.**

Radius (r) = \sqrt20

**Substitute any values you know into the equation.**

(x-4)^2+(y-(-8))^2=(\sqrt{20})^2

**Use the information you have to solve the problem.**

Simplify the equation.

\begin{aligned} & (x-4)^2+(y-(-8))^2=(\sqrt{20})^2 \\\\ & (x-4)^2+(y+8)^2=20 \end{aligned}

**Clearly state the answer.**

The equation of the circle is: (x-4)^2+(y+8)^2=20

- Be sure that students have a good understanding of coordinate geometry before asking them to graph circles and derive their equations or proofs.

- Let students use a graphing calculator to explore how changing the values for h, k and r change a circle.

**Forgetting the radius is squared in the equation**Remember, in the equation the radius is shown as a squared value. To find the radius, you need to square root this value.

**Thinking the radius can be negative**The radius cannot be negative because the radius is a length; it must always be a positive value.

**Confusing the location of the center with (x-h)^2+(y-k)^2=r^2**

The equation indicates how a circle would need to shift to return to the origin.

Therefore, the signs in the equation are opposite the signs in the coordinate.

For example,

An origin of (3, –4) would need to shift left 3 and up 4, which is

(x-3)^2+(y+4)^2=r^2.

**Confusing equation of a circle with quadratic equations**

The equation of a circle and a quadratic equation are not the same. A quadratic equation appears on a graph as a parabola, not a circle.

For example,

x^2+5x+4 is a quadratic equation.

This equation can be factored as (x+1)(x+4).

While they share some symbol commonalities, this is NOT the same as (x-h)^2+(y-k)^2=r^2 or x^2+y^2=r^2.

**Not recognizing the equation in other forms**

You may need to simplify, combine like terms or balance the sides of the equation to get the standard form of the equation of a circle.

- Linear graph

1) What is the equation of a circle with a radius of 4 and a center at the origin?

x^2+y^2=16

x^2+y^2=4

x^2+y^2=2

16

The equation of a circle with a center at the origin is x^2+y^2=r^2.

\begin{aligned} & x^2+y^2=r^2 \\\\ & x^2+y^2=4^2 \\\\ & x^2+y^2=16 \end{aligned}

2) What is the equation of a circle of the circle?

36

x^2+y^2=6

x^2+y^2=12

x^2+y^2=36

One diameter of the circle goes from (-6,0) to (6,0). This means the center is (0,0) or at the origin.

The equation of a circle with a center at the origin is x^2+y^2=r^2.

\begin{aligned} & x^2+y^2=r^2 \\\\ & x^2+y^2=6^2 \\\\ & x^2+y^2=36 \end{aligned}

3) What is the equation of a circle with a radius of \sqrt3 and a center at (-2, 3)?

(x-2)^2+(y-3)^2=1.73

(x+2)^2+(y-3)^2=9

(x-2)^2+(y-3)^2=3

(x+2)^2+(y-3)^2=3

The equation of a circle with a center at (h,k) is (x-h)^2+(y-k)^2=r^2.

\begin{aligned} & (x-h)^2+(y-k)^2=r^2 \\\\ & (x-(-2)^2+(y-3)^2=(\sqrt{3})^2 \\\\ & (x+2)^2+(y-3)^2=3 \end{aligned}

4) What is the radius of the circle with the equation x^2+y^2=100?

10,000

100

10

100,000

\begin{aligned} r^{2}&=100 \\\\ r&=\sqrt{100} \\\\ r&=10 \end{aligned}

x^2+y^2=10

5) What is the radius and the coordinates of the center of the circle with the equation \left(x-\cfrac{1}{2}\right)^2+(y+3)^2=25?

center: \left(-\cfrac{1}{2}, 3\right) and radius: 625

\left(\cfrac{1}{2},-3\right) and radius: 5

\left(-3, \cfrac{1}{2}\right) and radius: 625

1\left(-\cfrac{1}{2}, 3\right) and radius: 5

The equation of a circle with a center at (h,k) is

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

\begin{aligned} & \left(x-\cfrac{1}{2}\right)^2+(y+3)^2=25 \\\\ & \left(x-\cfrac{1}{2}\right)^2+(y-(-3))^2=5^2 \\\\ & h=\cfrac{1}{2} \\\\ & k=-3 \\\\ & r=5 \end{aligned}

6) What is the equation of the circle?

(x+1)^2+(y-3)^2=4

(x-1)^2+(y-3)^2=16

(x-1)^2+(y-3)^2=8

(x-1)^2+(y+3)^2=16

The equation of a circle with a center at (h,k) is (x-h)^2+(y-k)^2=r^2.

\begin{aligned} & (x-1)^2+(y-(-3))^2=4^2 \\\\ & (x-1)^2+(y+3)^2=16 \end{aligned}

The center is a fixed point inside the circle that is the same distance from all points on the circle. It can be found by finding the midpoint of any diameter.

Yes, but only if the same coefficient keeps the shape as a circle. Note also that to find the radius, the equation needs to be simplified.

For example, 3 x^2+3 y^2=9 will simplify to x^2+y^2=3.

They can be used to solve real world problems and problems in geometry (for example, finding a circular conic section).

- Number patterns
- Systems of equations
- Circle theorems
- Vectors
- Standard form

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[FREE] Common Core Practice Tests (Grades 3 to 6)

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