Circle Functions - Algebra 2

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 :

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)

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

The standard equation for a circle is:

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

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.

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.

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.

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:

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:

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:

The equation for a circle of radius centered on point has the equation

Just as with linear equations, the horizontal and vertical shifts are opposite of their sign, and are inside the parentheses. The equation of a circle simply must be memorized.

So the circle is centered on point , and plug those in and it yields the formula of the circle.

First determine the center of the circle. The "x-3" portion of the circle equation tells us that the x coordinate is equal to 3. The "y-3" portion of the circle equation tells us that the y coordinate is equal to 3 as well. Therefore, the center of the circle is at (3,3).

To find the distance between (3,3) and (0,0), it is necessary to use the Pythagorean Theorem . Where "a" and "b" are equal to 3

(to visualize, you may draw the two points on a graph, and create a triangle. The line connecting the two points is the hypotenuse, aka "c." )

Write the standard form for the equation of a circle.

The value of is and the value of is . The center of the circle is:

To find the radius, set and solve for .

Take the square root of both sides. We only consider the positive value since distance cannot be negative.

The answer is:

Recall that the standard equation of a circle is . Therefore, in this case . Square root both sides to find your radius. can be simplified to , which is your answer.

First, recall what the standard equation of a circle: . Your center is (h,k). Remember to flip the signs to get your center for this equation: .

Recall what the standard equation of a circle is:

.

is the center of the circle.

Remember that you have to change the signs!

Thus, since our equation is,

your answer for the center is: .

Step 1: Recall the general formula of a circle with a center that is not at .

, where is the center.

Step 2: Find the value of h and k here..

. To find this, we set of the general equation equal to (which is in the question)

. To find this, we set equal to .

The vertex is .

Step 3: To find the radius, take the square root of .

In the equation, we will get .

The radius of the circle is

The equation of a circle is in the format:

where is the center and is the radius.

Multiply two on both sides of the equation.

The equation becomes:

The center is .

The radius is .

The answer is:

To rewrite the given function as the equation of a circle in standard form, we must complete the square for x and y. This method requires us to use the following general form:

To start, we can complete the square for the x terms. We must halve the coefficient of x, square it, and add it to the first two terms:

Now, we can rewrite this as a perfect square, but because we added 4, we must subtract 4 as to not change the original function:

We do the same procedure for the y terms:

Rewriting our function, we get

Moving the constants to the right side, we get the function of a circle in standard form:

Comparing to

we see that the radius of the circle is

Notice that the radius is a distance and can therefore never be negative.