### All Algebra II Resources

## Example Questions

### Example Question #1 : Center And Radius Of Circle Functions

Consider a circle given by the formula:

.

This circle has a radius of ________ and is located at the point _________.

**Possible Answers:**

**Correct answer:**

The formula for a circle of radius , centered at the point is given by the general equation:

In this case, the radius is the square root of , which is six, and the center is at

### Example Question #69 : Quadratic Functions

What is the center and radius of the circle described by the equation:

**Possible Answers:**

(0,2); r=6

(0,-2); r=6

(0,-2); r=36

(0,2); r=36

**Correct answer:**

(0,-2); r=6

The standard equation for a circle is:

Therefore, the radius is 6 and the center is located at (0,-2)

### Example Question #70 : Quadratic Functions

Find the radius of the circle given by the equation:

**Possible Answers:**

**Correct answer:**

To find the center or the radius of a circle, first put the equation in the standard form for a circle: , where is the radius and is the center.

From our equation, we see that it has not yet been factored, so we must do that now. We can use the formula .

, so .

and , so and .

Therefore, .

Because the constant, in this case 4, was not in the original equation, we need to add it to both sides:

Now we do the same for :

We can now find :

### Example Question #71 : Quadratic Functions

Find the center of the circle given by the equation:

**Possible Answers:**

**Correct answer:**

To find the center or the radius of a circle, first put the equation in standard form: , where is the radius and is the center.

From our equation, we see that it has not yet been factored, so we must do that now. We can use the formula .

, so .

and , so and .

This gives .

Because the constant, in this case 9, was not in the original equation, we must add it to both sides:

Now we do the same for :

We can now find the center: (3, -9)

### Example Question #72 : Quadratic Functions

What is the center of the circular function ?

**Possible Answers:**

**Correct answer:**

Remember that the "shifts" involved with circular functions are sort of like those found in parabolas. When you shift a parabola left or right, you have to think "oppositely". A right shift requires you to subtract from the x-component, and a left one requires you to add. Hence, this circle has no horizontal shift, but does shift 6 upward for the vertical component.

You can also remember the general formula for a circle with center at and a radius of .

Comparing this to the given equation, we can determine the center point.

The center point is at (0,6) and the circle has a radius of 5.

### Example Question #73 : Quadratic Functions

What is the center of the circle described by ?

**Possible Answers:**

**Correct answer:**

Remember that the shifts for circles work in an opposite manner from what you might think. They are like the parabola's x-component. Hence, a subtracted variable actually means a shift up or to the right, for the vertical and horizontal components respectively. Since the x-component has a "+5", it is shifted left 5. Since the y-component has a , it is shifted upward 12. Therefore, this circle has a center at .

You can also remember the general formula for a circle with center at and a radius of .

Comparing this to the given equation, we can determine the center point.

The center point is at and the circle has a radius of 6.

### Example Question #74 : Quadratic Functions

What is the radius of the circle with equation ?

**Possible Answers:**

**Correct answer:**

Remember that for the equation of a circle, the lone number to the right of the equals sign is the radius squared.

The general formula for a circle with center at and a radius of is:

Comparing this to the given equation, we can determine the radius.

The center point is at and the circle has a radius of 9.

### Example Question #71 : Quadratic Functions

What is the sum of the values of the radius and center coordinates (both and ) for the given circle?

**Possible Answers:**

**Correct answer:**

Remember that the "shifts" involved with circular functions are sort of like those found in parabolas. When you shift a parabola left or right, you have to think "oppositely". A right shift requires you to subtract from the x-component, and a left one requires you to add. Hence, this circle has a positive 3 horizontal shift, and a negative 2 vertical shift.

You can also remember the general formula for a circle with center at and a radius of .

Comparing this to the given equation, we can determine the radius and center point.

The center point is at and the circle has a radius of 7.

The question asks us for the sum of these components:

### Example Question #76 : Quadratic Functions

What is the sum of the values of the radius and center coordinates (both and ) for the given circle?

**Possible Answers:**

**Correct answer:**

Remember that the "shifts" involved with circular functions are sort of like those found in parabolas. When you shift a parabola left or right, you have to think "oppositely". A right shift requires you to subtract from the x-component, and a left one requires you to add. Hence, this circle has a negative 5 horizontal shift, and a negative 22 vertical shift.

You can also remember the general formula for a circle with center at and a radius of .

Comparing this to the given equation, we can determine the radius and center point.

The center point is at and the circle has a radius of 11.

The question asks us for the sum of these components:

### Example Question #77 : Quadratic Functions

What is the sum of the values of the radius and center coordinates (both and ) for the given circle?

**Possible Answers:**

**Correct answer:**

Remember that the "shifts" involved with circular functions are sort of like those found in parabolas. When you shift a parabola left or right, you have to think "oppositely". A right shift requires you to subtract from the x-component, and a left one requires you to add. Hence, this circle has a positive 50 horizontal shift, and a negative 29 vertical shift.

You can also remember the general formula for a circle with center at and a radius of .

Comparing this to the given equation, we can determine the radius and center point.

The center point is at and the circle has a radius of 13.

The question asks us for the sum of these components:

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