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Algebra II : Quadratic Functions

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Example Question #981 :Algebra Ii

Determine the center and radius, respectively, given the equation:

x^2+y^2 +6x-2y-2 = 0

Possible Answers:

(-3,1); 144

$(-3,1); 2\sqrt3$

$(-3,1); 3\sqrt2$

(3,-1); 144

$(3,-1); 2\sqrt3$

Correct answer:

$(-3,1); 2\sqrt3$

Explanation:

In order to solve for the radius, we will need to complete the square twice.

Group the x and y-variables in parentheses. Starting from the original equation:

x^2+y^2 +6x-2y-2 = 0

(x^2+6x)+(y^2-2y)-2 = 0

Add two on both sides.

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

Divide by the second term coefficient of each binomial by 2, and add the squared quantity on both sides of the equation.

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

The equation becomes:

(x^2+6x+9)+(y^2-2y+1)=2+9+1

Factorize both polynomials in parentheses and simplify the right side.

(x+3)^2+(y-1)^2 =12

The center is: (h,k)= (-3,1)

The radius is: $r=\sqrt{12}=2\sqrt3$

The answer is: $(-3,1); 2\sqrt3$

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Example Question #1 :Graphing Circle Functions

Determine the graph of the equation

4(x-2)^2+4(y-3)^2=1

Possible Answers:

Ellipse, centered at (2,3)

Circle, centered at (2,3) with radius r= 1/2

Hyperbola, centered at (2,3)

Circle, centered at (2,3) with radius r= 1/4

Correct answer:

Circle, centered at (2,3) with radius r= 1/2

Explanation:

The equation of a circle in standard for is:

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

Where the center (x,y)=(h,k) and the radius of the cirlce is.

Dividing by 4 on both sides of the equation yields

$(x-2)^2+(y-3)^2 = \frac{1}{4}$

$(x-2)^2+(y-3)^2 = \bigg(\frac{1}{2}\bigg)^2$

an equation whose graph is a circle, centered at (2,3) with radius = .5

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Example Question #2 :Graphing Circle Functions

Give the radius and the center of the circle for the equation below.

$(x-3)^{2}+(y + 4)^2 = 16$

Possible Answers:

r = 16

$\text{Center} = (3, -4)$

r = 16

$\text{Center} = (3, 4)$

r = 4

$\text{Center} = (-3, 4)$

r = 4

$\text{Center} = (3, 4)$

r = 4

$\text{Center} = (3, -4)$

Correct answer:

r = 4

$\text{Center} = (3, -4)$

Explanation:

Look at the formula for the equation of a circle below.

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

Here (h, k) is the center andis the radius. Notice that the subtraction in the center is part of the formula. Thus, looking at our equation it is clear that the center is (3, -4) and the radius squared is. When we square root this value we get that the radius must be.

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Example Question #1 :Graphing Circle Functions

Determine the equation of a circle whose center lies at the point $\left ( 3,4\right )$ and has a radius of.

Possible Answers:

(x+3)^2+(y+4)^2=1

$\frac{x^2}{3}+\frac{y^2}{4}-1=0$

$\frac{x^2}{3}-\frac{y^2}{4}=1$

(x-3)^2+(y-4)^2=1

Correct answer:

(x-3)^2+(y-4)^2=1

Explanation:

The equation for a circle with center (h,k) and radiusis :

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

Our circle is centered at (3,4) with radius r=1 , so the equation for this circle is :

(x-3)^2+(y-4)^2=1

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Example Question #4 :Graphing Circle Functions

$\small f(x)=(x-2)^2+(y+5)^2=36$

What is the radius of the circle?

Possible Answers:

$\small \small r=2$

$\small r=6$

$\small \small r=36$

$\small \small r=3$

$\small \small r=5$

Correct answer:

$\small r=6$

Explanation:

The parent equation of a circle is represented by $\small f(x)=(x-h)^2+(y-k)^2=r^2$ . The radius of the circle is equal to $\small r$ . The radius of the cirle is $\small \sqrt{36}$ .

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Example Question #5 :Graphing Circle Functions

What is the center of the circle expressed by the funciton $\small x^2+4x+y^2=5$ ?

Possible Answers:

$\small (2,0)$

$\small (-2,0)$

$\small (0,0)$

$\small (5,0)$

$\small (-5,0)$

Correct answer:

$\small (-2,0)$

Explanation:

The equation can be rewritten so that it looks like the parent equation for a circle $\small (x-h)^2+(y-k)^2=r^2$ . After completeing the square, the equation changes from $\small x^2+4x+y^2=5$ to $\small x^2+4x+4+y^2=9$ . From there it can be expressed as $\small (x+2)^2+y^2=9$ . Therefore the center of the circle is at $\small (-2,0)$ .

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Example Question #6 :Graphing Circle Functions

The graph of the equation

$x^{2} + y^{2} + 10x - 8y = 84$

is a circle with what as the length of its radius?

Possible Answers:

$10\sqrt{5}$

$5\sqrt{5}$

Correct answer:

$5\sqrt{5}$

Explanation:

Rewrite the equation of the circle in standard form

$(x - h)^{2} + \left ( y - k\right )^{2} = r^{2}$

as follows:

$x^{2} + y^{2} + 10x - 8y = 84$

$x^{2} + 10x+ y^{2} - 8y = 84$

Since $\left (\frac{10}{2} \right ) ^{2} = 5 ^{2} = 25$ and $\left (\frac{-8}{2} \right ) ^{2} = (-4)^{2} = 16$ , we complete the squares by adding:

$x^{2} + 10x+ 25 + y^{2} - 8y + 16 = 84 + 25 + 16$

$(x+5) ^{2} + \left (y- 4 \right ) ^{2} = 125$

The standard form of the equation sets

$r^{2} = 125$ ,

so the radius of the circle is

$r = \sqrt{125} = \sqrt{25} \cdot \sqrt{5} = 5 \sqrt{5}$

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Example Question #7 :Graphing Circle Functions

What are the coordinates of the center of a circle with the equation (x - 1)^2 + (y - 6)^2 = 25 ?

Possible Answers:

(-6, -1)

(1, 6)

(-1, -6)

(6, 1)

Correct answer:

(1, 6)

Explanation:

The equation of a circle is (x - h)^2 + (y - k)^2 = r^2 , in which (h, k) is the center of the circle. To derive the center of a circle from its equation, identify the constants immediately following x and y, and flip their signs. In the given equation, x is followed by -1 and y is followed by -6, so the coordinates of the center must be (1, 6).

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Example Question #8 :Graphing Circle Functions

What is the radius of a circle with the equation x^2 + 6x + y^2 + 2y = -6 ?

Possible Answers:

Correct answer:

Explanation:

To convert the given equation into the format (x - h)^2 + (y - k)^2 = r^2 , complete the square by adding (b/2a)^2 to the x-terms and to the y-terms.

x^2 + 6x + y^2 + 2y = -6

x^2 + 6x + (6/2)^2 + y^2 + 2y + (2/2)^2 = -6 + (6/2)^2 + (2/2)^2

x^2 + 6x + 9 + y^2 + 2y + 1 = -6 + 9 + 1

(x + 3)^2 + (y + 1)^2 = 4

The square root of 4 is 2, so the radius of the circle is 2.

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Example Question #9 :Graphing Circle Functions

What is the radius of a circle with the equation x^2 + 8x + y^2 - 4y = 5 ?

Possible Answers:

Correct answer:

Explanation:

To convert the given equation into the format (x - h)^2 + (y - k)^2 = r^2 , complete the square by adding (b/2a)^2 to the x-terms and to the y-terms.

x^2 + 8x + y^2 - 4y = 5

x^2 + 8x + (8/2)^2 + y^2 - 4y + (-4/2)^2 = 5 + (8/2)^2 + (-4/2)^2

x^2 + 8x + 16 + y^2 - 4y + 4 = 5 + 16 + 4

(x + 4)^2) + (y - 2)^2 = 25

The square root of 25 is 5, so the radius of the circle is 5.

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Algebra II : Quadratic Functions

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #981 :Algebra Ii

Example Question #1 :Graphing Circle Functions

Example Question #2 :Graphing Circle Functions

Example Question #1 :Graphing Circle Functions

Example Question #4 :Graphing Circle Functions

Example Question #5 :Graphing Circle Functions

Example Question #6 :Graphing Circle Functions

Example Question #7 :Graphing Circle Functions

Example Question #8 :Graphing Circle Functions

Example Question #9 :Graphing Circle Functions

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