Graphing Circle Functions - Algebra 2

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The graph of the equation

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

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Rewrite the equation of the circle in standard form

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

as follows:

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:

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

The standard form of the equation sets

,

so the radius of the circle is

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

Determine the graph of the equation

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The equation of a circle in standard for is:

Where the center and the radius of the cirlce is .

Dividing by 4 on both sides of the equation yields

or

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

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

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

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

Here is the center and is 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 and the radius squared is . When we square root this value we get that the radius must be .

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

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The equation for a circle with center and radius is :

Our circle is centered at with radius , so the equation for this circle is :

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

What is the radius of the circle?

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

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

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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 $\small (-2,0)$ .

What are the coordinates of the center of a circle with the equation ?

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The equation of a circle is , 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).

What is the radius of a circle with the equation ?

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To convert the given equation into the format , complete the square by adding to the x-terms and to the y-terms.

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

What is the radius of a circle with the equation ?

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To convert the given equation into the format , complete the square by adding to the x-terms and to the y-terms.

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

Which equation does this graph represent?

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The equation of a circle is , in which (h, k) is the center of the circle and r is its radius. Because the graph of the circle is centered at (0, 0), h and k are both 0. Because the radius is 3, the right side of the equation is equal to 9.

Which equation does this graph represent?

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The equation of a circle is , in which (h, k) is the center of the circle and r is its radius. Because the graph of the circle is centered at (2, -3), h and k are -2 and 3. Because the radius is 4, the right side of the equation is equal to 16.

The graph of the equation

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

Tap to see back →

Rewrite the equation of the circle in standard form

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

as follows:

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:

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

The standard form of the equation sets

,

so the radius of the circle is

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

Determine the graph of the equation

Tap to see back →

The equation of a circle in standard for is:

Where the center and the radius of the cirlce is .

Dividing by 4 on both sides of the equation yields

or

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

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

Tap to see back →

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

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

Here is the center and is 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 and the radius squared is . When we square root this value we get that the radius must be .

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

Tap to see back →

The equation for a circle with center and radius is :

Our circle is centered at with radius , so the equation for this circle is :

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

What is the radius of the circle?

Tap to see back →

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

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

Tap to see back →

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 $\small (-2,0)$ .

What are the coordinates of the center of a circle with the equation ?

Tap to see back →

The equation of a circle is , 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).

What is the radius of a circle with the equation ?

Tap to see back →

To convert the given equation into the format , complete the square by adding to the x-terms and to the y-terms.

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

What is the radius of a circle with the equation ?

Tap to see back →

To convert the given equation into the format , complete the square by adding to the x-terms and to the y-terms.

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