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Algebra II : Quadratic Functions

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Example Questions

Example Question #7 : 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+5)^2+(y+22)^2=121

Possible Answers:

148

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 negative 5 horizontal shift, and a negative 22 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+5)^2+(y+22)^2=121

$h=-5,\ k=-22,\ r=11$

The center point is at (-5,-22) and the circle has a radius of 11.

The question asks us for the sum of these components:

(-5)+(-22)+11=-16

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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 #81 : Quadratic Functions

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

Possible Answers:

(4,-2)

(7,7)

(-4,2)

(4,2)

(-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 #82 : Quadratic Functions

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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Algebra II : Quadratic Functions

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #7 : Center And Radius Of Circle Functions

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 #81 : Quadratic Functions

Example Question #12 : Center And Radius Of Circle Functions

Example Question #82 : Quadratic Functions

Example Question #18 : Center And Radius Of Circle Functions

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Algebra II : Quadratic Functions

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #7 : Center And Radius Of Circle Functions

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 #81 : Quadratic Functions

Example Question #12 : Center And Radius Of Circle Functions

Example Question #82 : Quadratic Functions

Example Question #18 : Center And Radius Of Circle Functions

All Algebra II Resources

Report an issue with this question

DMCA Complaint

Contact Information

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