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Common Core: High School - Geometry : Expressing Geometric Properties with Equations

Study concepts, example questions & explanations for Common Core: High School - Geometry

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All Common Core: High School - Geometry Resources

6 Diagnostic Tests 114 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Derive Circle Equation: Ccss.Math.Content.Hsg Gpe.A.1

What is the equation of the circle shown below?
Plot1

Possible Answers:

$\left(x + 7\right)^{2} + \left(y + 2\right)^{2} = 4$

$\left(x - 7\right)^{2} + \left(y - 2\right)^{2} = 16$

$\left(x + 7\right)^{2} + \left(y + 2\right)^{2} = 16$

$\left(x + 7\right)^{2} + \left(y - 2\right)^{2} = 4$

$\left(x - 7\right)^{2} + \left(y + 2\right)^{2} = 16$

Correct answer:

$\left(x + 7\right)^{2} + \left(y + 2\right)^{2} = 16$

Explanation:

In order to find the equation, we must find the coordinates of the center of the circle.

If we look at the picture,

Plot1

we can see that the center is at

$\left ( -7, -2\right )$

The next step it to find the radius. Recall the radius is the distance from the center of the circle to any point on the edge of the circle.

From looking at the picture, we can see that the radius is 4. With this information, we can plug it into the general circle equation

The general circle equation is,

$\left(- a + x\right)^{2} + \left(- b + y\right)^{2} = c^{2}$

Now we substitute for $\uptext{a}$ , $\uptext{b}$ , and $\uptext{c}$ . When we plug in the values, we get

$\left(x + 7\right)^{2} + \left(y + 2\right)^{2} = 16$

Report an Error

Example Question #2 : Derive Circle Equation: Ccss.Math.Content.Hsg Gpe.A.1

What is the equation of the circle shown below?
Plot2

Possible Answers:

$\left(x - 9\right)^{2} + \left(y - 5\right)^{2} = 36$

$\left(x + 9\right)^{2} + \left(y + 5\right)^{2} = 36$

$\left(x - 9\right)^{2} + \left(y + 5\right)^{2} = 36$

$\left(x + 9\right)^{2} + \left(y + 5\right)^{2} = 6$

$\left(x + 9\right)^{2} + \left(y - 5\right)^{2} = 6$

Correct answer:

$\left(x + 9\right)^{2} + \left(y + 5\right)^{2} = 36$

Explanation:

In order to find the equation, we must find the coordinates of the center of the circle.

If we look at the picture,

Plot2

we can see that the center is at $\left ( -9, -5\right )$ .

The next step it to find the radius. Recall the radius is the distance from the center of the circle to any point of the circle's edge.

From looking at the picture, we can see that the radius is 6.

With this information, we can plug it into the general circle equation.

The general circle equation is

$\left(- a + x\right)^{2} + \left(- b + y\right)^{2} = c^{2}$

Now we substitute for $\uptext{a}$ , $\uptext{b}$ , and $\uptext{c}$ .

When we plug in the values, we get

$\left(x + 9\right)^{2} + \left(y + 5\right)^{2} = 36$

Report an Error

Example Question #3 : Derive Circle Equation: Ccss.Math.Content.Hsg Gpe.A.1

What is the equation of the circle shown below?

Plot3

Possible Answers:

$\left(x - 9\right)^{2} + \left(y - 2\right)^{2} = 36$

$\left(x + 9\right)^{2} + \left(y - 2\right)^{2} = 6$

Correct answer:

Explanation:

In order to find the equation, we must find the coordinates of the center of the circle.

If we look at the picture,

Plot3

we can see that the center is at $\left ( -9, -2\right )$ .

The next step it to find the radius From looking at the picture, we can see that the radius is 6.

With this information, we can plug it into the general circle equation.

The general circle equation is

$\left(- a + x\right)^{2} + \left(- b + y\right)^{2} = c^{2}$

Now we substitute for $\uptext{a}$ , $\uptext{b}$ , and $\uptext{c}$ .

We plug in the values, we get

$\left(x + 9\right)^{2} + \left(y + 2\right)^{2} = 36$

Report an Error

Example Question #4 : Expressing Geometric Properties With Equations

What is the equation of the circle shown below?

Plot4

Possible Answers:

$\left(x - 10\right)^{2} + \left(y + 4\right)^{2} = 16$

$\left(x + 10\right)^{2} + \left(y + 4\right)^{2} = 4$

$\left(x + 10\right)^{2} + \left(y - 4\right)^{2} = 16$

$\left(x + 10\right)^{2} + \left(y - 4\right)^{2} = 4$

$\left(x - 10\right)^{2} + \left(y - 4\right)^{2} = 16$

Correct answer:

$\left(x + 10\right)^{2} + \left(y - 4\right)^{2} = 16$

Explanation:

In order to find the equation, we must find the coordinates of the center of the circle.

If we look at the picture,

Plot4

we can see that the center is at $\left ( -10, 4\right )$

The next step it to find the radius From looking at the picture, we can see that the radius is 4.

With this information, we can plug it into the general circle equation.

The general circle equation is

$\left(- a + x\right)^{2} + \left(- b + y\right)^{2} = c^{2}$

Now we substitute for $\uptext{a}$ , $\uptext{b}$ , and $\uptext{c}$

We plug in the values, we get

$\left(x + 10\right)^{2} + \left(y - 4\right)^{2} = 16$

Report an Error

Example Question #5 : Expressing Geometric Properties With Equations

What is the equation of the circle shown below?
Plot5

Possible Answers:

$\left(x - 4\right)^{2} + \left(y - 2\right)^{2} = 81$

$\left(x + 4\right)^{2} + \left(y + 2\right)^{2} = 81$

$\left(x + 4\right)^{2} + \left(y - 2\right)^{2} = 9$

$\left(x + 4\right)^{2} + \left(y + 2\right)^{2} = 9$

$\left(x - 4\right)^{2} + \left(y + 2\right)^{2} = 81$

Correct answer:

$\left(x + 4\right)^{2} + \left(y + 2\right)^{2} = 81$

Explanation:

In order to find the equation, we must find the coordinates of the center of the circle.

If we look at the picture, we can see that the center is at $\left ( -4, -2\right )$

The next step it to find the radius. From looking at the picture,

Plot5

we can see that the radius is 9.

With this information, we can plug it into the general circle equation.

The general circle equation is

$\left(- a + x\right)^{2} + \left(- b + y\right)^{2} = c^{2}$

Now we substitute for $\uptext{a}$ , $\uptext{b}$ , and $\uptext{c}$ .

We plug in the values, we get

$\left(x + 4\right)^{2} + \left(y + 2\right)^{2} = 81$

Report an Error

Example Question #6 : Expressing Geometric Properties With Equations

What is the equation of the circle shown below?
Plot6

Possible Answers:

$\left(x + 3\right)^{2} + \left(y - 3\right)^{2} = 1$

$\left(x - 3\right)^{2} + \left(y + 3\right)^{2} = 1$

$\left(x + 3\right)^{2} + \left(y + 3\right)^{2} = 1$

$\left(x - 3\right)^{2} + \left(y - 3\right)^{2} = 1$

Correct answer:

$\left(x + 3\right)^{2} + \left(y - 3\right)^{2} = 1$

Explanation:

In order to find the equation, we must find the coordinates of the center of the circle.

If we look at the picture,

Plot6

we can see that the center is at $\left ( -3, 3\right )$

The next step it to find the radius From looking at the picture, we can see that the radius is 1.

With this information, we can plug it into the general circle equation.

The general circle equation is

$\left(- a + x\right)^{2} + \left(- b + y\right)^{2} = c^{2}$

Now we substitute for $\uptext{a}$ , $\uptext{b}$ , and $\uptext{c}$

We plug in the values, we get

$\left(x + 3\right)^{2} + \left(y - 3\right)^{2} = 1$

Report an Error

Example Question #7 : Expressing Geometric Properties With Equations

What is the equation of the circle shown below?
Plot7

Possible Answers:

Correct answer:

Explanation:

In order to find the equation, we must find the coordinates of the center of the circle.

If we look at the picture,

Plot7

we can see that the center is at $\left ( 2, 2\right )$

The next step it to find the radius From looking at the picture, we can see that the radius is 4.

With this information, we can plug it into the general circle equation.

The general circle equation is

$\left(- a + x\right)^{2} + \left(- b + y\right)^{2} = c^{2}$

Now we substitute for $\uptext{a}$ , $\uptext{b}$ , and $\uptext{c}$

We plug in the values, we get

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

Report an Error

Example Question #8 : Expressing Geometric Properties With Equations

What is the equation of the circle shown below?

Plot8

Possible Answers:

$\left(x + 10\right)^{2} + \left(y + 3\right)^{2} = 6$

$\left(x + 10\right)^{2} + \left(y - 3\right)^{2} = 6$

$\left(x - 10\right)^{2} + \left(y - 3\right)^{2} = 36$

$\left(x - 10\right)^{2} + \left(y + 3\right)^{2} = 36$

$\left(x + 10\right)^{2} + \left(y + 3\right)^{2} = 36$

Correct answer:

$\left(x + 10\right)^{2} + \left(y + 3\right)^{2} = 36$

Explanation:

In order to find the equation, we must find the coordinates of the center of the circle If we look at the picture,

Plot8

we can see that the center is at $\left ( -10, -3\right )$

The next step it to find the radius. From looking at the picture, we can see that the radius is 6. With this information, we can plug it into the general circle equation.

The general circle equation is

$\left(- a + x\right)^{2} + \left(- b + y\right)^{2} = c^{2}$

Now we substitute for $\uptext{a}$ , $\uptext{b}$ , and $\uptext{c}$ .

We plug in the values, we get

$\left(x + 10\right)^{2} + \left(y + 3\right)^{2} = 36$

Report an Error

Example Question #9 : Expressing Geometric Properties With Equations

What is the equation of the circle shown below?

Plot9

Possible Answers:

$\left(x - 1\right)^{2} + \left(y - 7\right)^{2} = 2$

$\left(x + 1\right)^{2} + \left(y + 7\right)^{2} = 4$

$\left(x + 1\right)^{2} + \left(y - 7\right)^{2} = 4$

$\left(x - 1\right)^{2} + \left(y + 7\right)^{2} = 2$

$\left(x - 1\right)^{2} + \left(y - 7\right)^{2} = 4$

Correct answer:

$\left(x - 1\right)^{2} + \left(y - 7\right)^{2} = 4$

Explanation:

In order to find the equation, we must find the coordinates of the center of the circle.

If we look at the picture,

Plot9

we can see that the center is at $\left ( 1, 7\right )$

The next step it to find the radius. From looking at the picture, we can see that the radius is 2. With this information, we can plug it into the general circle equation.

The general circle equation is

$\left(- a + x\right)^{2} + \left(- b + y\right)^{2} = c^{2}$

Now we substitute for $\uptext{a}$ , $\uptext{b}$ , and $\uptext{c}$

We plug in the values, we get

$\left(x - 1\right)^{2} + \left(y - 7\right)^{2} = 4$

Report an Error

Example Question #10 : Expressing Geometric Properties With Equations

What is the equation of the circle shown below?

Plot10

Possible Answers:

$\left(x + 5\right)^{2} + \left(y + 6\right)^{2} = 2$

$\left(x - 5\right)^{2} + \left(y - 6\right)^{2} = 4$

$\left(x - 5\right)^{2} + \left(y + 6\right)^{2} = 4$

Correct answer:

Explanation:

In order to find the equation, we must find the coordinates of the center of the circle.

If we look at the picture,
Plot10
we can see that the center is at $\left ( -5, 6\right )$

The next step it to find the radius. From looking at the picture, we can see that the radius is 2. With this information, we can plug it into the general circle equation.

The general circle equation is

$\left(- a + x\right)^{2} + \left(- b + y\right)^{2} = c^{2}$

Now we substitute for $\uptext{a}$ , $\uptext{b}$ , and $\uptext{c}$

We plug in the values, we get

$\left(x + 5\right)^{2} + \left(y - 6\right)^{2} = 4$

Report an Error

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All Common Core: High School - Geometry Resources

6 Diagnostic Tests 114 Practice Tests Question of the Day Flashcards Learn by Concept

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Common Core: High School - Geometry : Expressing Geometric Properties with Equations

Study concepts, example questions & explanations for Common Core: High School - Geometry

All Common Core: High School - Geometry Resources

Example Questions

Example Question #1 : Derive Circle Equation: Ccss.Math.Content.Hsg Gpe.A.1

Example Question #2 : Derive Circle Equation: Ccss.Math.Content.Hsg Gpe.A.1

Example Question #3 : Derive Circle Equation: Ccss.Math.Content.Hsg Gpe.A.1

Example Question #4 : Expressing Geometric Properties With Equations

Example Question #5 : Expressing Geometric Properties With Equations

Example Question #6 : Expressing Geometric Properties With Equations

Example Question #7 : Expressing Geometric Properties With Equations

Example Question #8 : Expressing Geometric Properties With Equations

Example Question #9 : Expressing Geometric Properties With Equations

Example Question #10 : Expressing Geometric Properties With Equations

All Common Core: High School - Geometry Resources

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