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Algebra II : Functions and Graphs

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Example Questions

Example Question #8 : Center And Radius Of Circle Functions

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

(x-50)^2+(y+29)^2=169

Possible Answers:

148

190

Correct answer:

Explanation:

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 {(h,k)} and a radius of .

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

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

(x-50)^2+(y+29)^2=169

$h=50,\ k=-29,\ r=13$

The center point is at (50,-29) and the circle has a radius of 13.

The question asks us for the sum of these components:

50+(-29)+13=34

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Example Question #81 : Quadratic Functions

Screen_shot_2014-12-24_at_4.27.17_pm

What is the equation of the above circle?

Possible Answers:

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

x^2-2+y^2+1=4

x^2+y^2-4=0

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

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

Correct answer:

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

Explanation:

The equation for a circle of radius centered on point (a,b) has the equation (x-a)^2+(y-b)^2=r^2

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 (2,-1) , and plug those in and it yields the formula of the circle.

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Example Question #11 : Center And Radius Of Circle Functions

A circle is graphed by the equation (x-3) + (y-3) = 9. What is the distance from the center of the circle to the point (0,0) on a standard coordinate plane?

Possible Answers:

$\sqrt{3}$

$3 \sqrt{2}$

$\frac{2}{3}$

$2\sqrt{3}$

Correct answer:

$3 \sqrt{2}$

Explanation:

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 $a^{2} + b^{2}= c^{2}$ . 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." )

$3^{2}+ 3^{2}= 18$

$\sqrt{18} = 3\sqrt{2}$

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Example Question #81 : Quadratic Functions

Find the center and radius for the equation: (x+1)^2 + (y-4)^2 = 6

Possible Answers:

$\textup{Center: }(1,-4) \textup{ Radius: } \sqrt{6}$

$\textup{Center: }(-1,4) \textup{ Radius: } \sqrt{6}$

$\textup{Center: }(-1,4) \textup{ Radius: } 2\sqrt{3}$

$\textup{Center: }(1,-4) \textup{ Radius: } 36$

$\textup{Center: }(-1,4) \textup{ Radius: } 36$

Correct answer:

$\textup{Center: }(-1,4) \textup{ Radius: } \sqrt{6}$

Explanation:

Write the standard form for the equation of a circle.

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

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

(h,k)=(-1,4)

To find the radius, set r^2=6 and solve for .

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

$\sqrt{r^2}=\sqrt{6}$

$r=\sqrt6$

The answer is: $\textup{Center: }(-1,4) \textup{ Radius: } \sqrt{6}$

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Example Question #14 : Center And Radius Of Circle Functions

What is the radius of a circle with the equation x^2+y^2=24 ?

Possible Answers:

$6\sqrt{2}$

$2\sqrt{6}$

Correct answer:

$2\sqrt{6}$

Explanation:

Recall that the standard equation of a circle is x^2+y^2=r^2 . Therefore, in this case r^2=24 . Square root both sides to find your radius. $\sqrt{24}$ can be simplified to $2\sqrt{6}$ , which is your answer.

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Example Question #481 : Functions And Graphs

What is the center of this circle: (x-4)^2+(y+2)^2=49 ?

Possible Answers:

(-4,-2)

(4,-2)

(-4,2)

(7,7)

(4,2)

Correct answer:

(4,-2)

Explanation:

First, recall what the standard equation of a circle: (x-h)^2+(y-k)^2=r^2 . Your center is (h,k). Remember to flip the signs to get your center for this equation: (4,-2) .

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Example Question #12 : Center And Radius Of Circle Functions

What is the center of this circle: (x-4)^2+(y+9)^2=18 ?

Possible Answers:

(4, 9)

(4,-9)

(-4,-9)

(-4,9)

Correct answer:

(4,-9)

Explanation:

Recall what the standard equation of a circle is:

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

(h,k) is the center of the circle.

Remember that you have to change the signs!

Thus, since our equation is,

(x-4)^2+(y+9)^2=18

your answer for the center is: (4,-9) .

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Example Question #482 : Functions And Graphs

What is the radius of this circle: x^2+(y-4)^2=49 ?

Possible Answers:

Correct answer:

Explanation:

Recall that the standard equation of a circle is

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

Therefore, looking at the equation given,

x^2+(y-4)^2=49

$\\r^2=49 \\\sqrt{r^2}=\sqrt{49} \\r=7$ .

Solve for r to get 7.

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Example Question #18 : Center And Radius Of Circle Functions

What is the center and radius of a circle with an equation: (x-4)^2+(y+9)^2=25 ?

Possible Answers:

Center=(-4,-9), Radius=5

None of the Above

Center=(-4,9), Radius=5

Center=(4,-9), Radius=5

Correct answer:

Center=(4,-9), Radius=5

Explanation:

Step 1: Recall the general formula of a circle with a center that is not at (0,0) .

(x-h)^2+(y-k)^2=r^2 , where (h,k) is the center.

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

h=4 . To find this, we set of the general equation equal to (which is in the question)

k=-9 . To find this, we set equal to .

The vertex is (4,-9) .

Step 3: To find the radius, take the square root of r^2 .

In the equation, we will get $\sqrt{25}=5$ .

The radius of the circle is

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Example Question #481 : Functions And Graphs

Find the center and radius of the circle given the following equation:

$\frac{(x-3)^2}{2}+\frac{(y-6)^2}{2} = 4$

Possible Answers:

c= (3,6), r= 2

$c= (-3,-6), r= 2\sqrt2$

$\textup{The equation does not resemble a circle.}$

$c= (3,6), r= 2\sqrt2$

c= (-3,-6), r= 2

Correct answer:

$c= (3,6), r= 2\sqrt2$

Explanation:

The equation of a circle is in the format:

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

where (h,k) is the center and is the radius.

Multiply two on both sides of the equation.

$2[\frac{(x-3)^2}{2}+\frac{(y-6)^2}{2} = 4]$

The equation becomes:

(x-3)^2+(y-6)^2=8

The center is (3,6) .

The radius is $\sqrt8 = 2\sqrt2$ .

The answer is: $c= (3,6), r= 2\sqrt2$

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Algebra II : Functions and Graphs

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #8 : Center And Radius Of Circle Functions

Example Question #81 : Quadratic Functions

Example Question #11 : Center And Radius Of Circle Functions

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 $a^{2} + b^{2}= c^{2}$ . 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." )

Example Question #81 : Quadratic Functions

Example Question #14 : Center And Radius Of Circle Functions

Example Question #481 : Functions And Graphs

Example Question #12 : Center And Radius Of Circle Functions

Example Question #482 : Functions And Graphs

Example Question #18 : Center And Radius Of Circle Functions

Example Question #481 : Functions And Graphs

All Algebra II Resources

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Algebra II : Functions and Graphs

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #8 : Center And Radius Of Circle Functions

Example Question #81 : Quadratic Functions

Example Question #11 : Center And Radius Of Circle Functions

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." )

Example Question #81 : Quadratic Functions

Example Question #14 : Center And Radius Of Circle Functions

Example Question #481 : Functions And Graphs

Example Question #12 : Center And Radius Of Circle Functions

Example Question #482 : Functions And Graphs

Example Question #18 : Center And Radius Of Circle Functions

Example Question #481 : Functions And Graphs

All Algebra II Resources

Report an issue with this question

DMCA Complaint

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Address

Complaint Details

To find the distance between (3,3) and (0,0), it is necessary to use the Pythagorean Theorem $a^{2} + b^{2}= c^{2}$ . Where "a" and "b" are equal to 3