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Algebra II : Circle Functions

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Example Question #41 : Circle Functions

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

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

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

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

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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Example Question #42 : Circle Functions

Determine the radius of the circle given by the following function:

x^2-4x+y^2-2x=11

Possible Answers:

Correct answer:

Explanation:

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:

(a+b)^2=a^2+2ab+b^2

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:

(-2)^2=4

x^2-4x+4

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:

(x-2)^2-4

We do the same procedure for the y terms:

(-1)^2=1

y^2-2x+1

(y-1)^2-1

Rewriting our function, we get

(x-2)^2-4+(y-1)^2-1=11

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

(x-2)^2+(y-1)^2=16

Comparing to

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

we see that the radius of the circle is

$\sqrt{16}=4$

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

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Example Question #43 : Circle Functions

$Identify\ the\ center\ and\ radius\ for\ (x+4)^2+(y-2)^2=49$

Possible Answers:

Center: (-4,2); Radius= 49

Center: (4,-2); Radius= 7

Center: (-4,2); Radius= 7

Center: (4,-2); Radius= 49

Correct answer:

Center: (-4,2); Radius= 7

Explanation:

When identifying the center of a circle, take the opposite sign of each value connected to x and y.

$(x+4)^2+(y-2)^2=49\ compare\ with\ the\ general\ equation\ of\ a\ circle: (x-h)^2+(y-k)^2=r^2$

$Changing\ signs\ for\ each\ gives\ us\ a\ center\ of: (-4,2)$

$To\ find\ the\ radius\ we\ set\ r^2=49,$

$take\ square\ root\ of\ both\ sides\ to\ get\ r=7.$

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Example Question #44 : Circle Functions

What is the center and radius of the following equation, respectively?

(x-6)^2+(y+7)^2 = 2

Possible Answers:

$(6,-7); \sqrt2$

(6,-7); 2

(6,-7);4

(-6,7); 2

$(-6,7); \sqrt2$

Correct answer:

$(6,-7); \sqrt2$

Explanation:

The equation given represents a circle.

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

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

h=6 , k = -7

The center is at: (6,-7)

Set up an equation to solve the radius.

r^2=2

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

The answer is: $(6,-7); \sqrt2$

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Example Question #45 : Circle Functions

Which of the following represents the formula of a circle with a radius of centered at (-1,2) ?

Possible Answers:

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

$(x+1)^2+(y-2)^2=\sqrt2$

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

$(x-1)^2+(y+2)^2=\sqrt2$

$(x-1)^2+(y+2)^2=2\sqrt2$

Correct answer:

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

Explanation:

Write the standard form for a circle.

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

The circle is centered at: (h,k)

The radius is: r=2

Substitute all the known values into the formula.

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

Simplify this equation.

The answer is: (x+1)^2+(y-2)^2=4

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Algebra II : Circle Functions

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #41 : Circle Functions

Example Question #42 : Circle Functions

Example Question #43 : Circle Functions

Example Question #44 : Circle Functions

Example Question #45 : Circle Functions

All Algebra II Resources

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