Circle Functions

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Math › Circle Functions

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Find the center and radius of the circle defined by the equation:

$\left ( 3,-5 \right )\ and\ r=6$

$\left ( 3,5 \right )\ and\ r=6$

$\left ( -3,-5 \right )\ and\ r=18$

$\left ( -3,5 \right )\ and\ r=18$

$\left ( -3,5 \right )\ and\ r=6$

Explanation

The equation of a circle is: $(x-x_{1})^2+(y-y_{1})^2=r^2$ where is the radius and $(x_{1},y_{1})$ is the center.

In this problem, the equation is already in the format required to determine center and radius. To find the -coordinate of the center, we must find the value of that makes equal to 0, which is 3. We do the same to find the y-coordinate of the center and find that . To find the radius we take the square root of the constant on the right side of the equation which is 6.

Find the center and radius of the circle defined by the equation:

$\left ( 3,-5 \right )\ and\ r=6$

$\left ( 3,5 \right )\ and\ r=6$

$\left ( -3,-5 \right )\ and\ r=18$

$\left ( -3,5 \right )\ and\ r=18$

$\left ( -3,5 \right )\ and\ r=6$

Explanation

The equation of a circle is: $(x-x_{1})^2+(y-y_{1})^2=r^2$ where is the radius and $(x_{1},y_{1})$ is the center.

Find the center and radius of the circle defined by the equation:

$(x+12)^{2}+(y+10)^2=100$

$\left ( -12,-10 \right )\ and\ r=10$

$\left ( 10,-12 \right )\ and\ r=100$

$\left ( 12,-10 \right )\ and\ r=10$

$\left ( -12,10 \right )\ and\ r=25$

$\left ( 12,-10 \right )\ and\ r=100$

Explanation

The equation of a circle is: $(x-x_{1})^2+(y-y_{1})^2=r^2$ where is the radius and $(x_{1},y_{1})$ is the center.

In this problem, the equation is already in the format required to determine center and radius. To find the -coordinate of the center, we must find the value of that makes equal to , which is . We do the same to find the y-coordinate of the center and find that . To find the radius we take the square root of the constant on the right side of the equation which is 10.

Find the center and radius of the circle defined by the equation:

$(x+12)^{2}+(y+10)^2=100$

$\left ( -12,-10 \right )\ and\ r=10$

$\left ( 10,-12 \right )\ and\ r=100$

$\left ( 12,-10 \right )\ and\ r=10$

$\left ( -12,10 \right )\ and\ r=25$

$\left ( 12,-10 \right )\ and\ r=100$

Explanation

The equation of a circle is: $(x-x_{1})^2+(y-y_{1})^2=r^2$ where is the radius and $(x_{1},y_{1})$ is the center.

Find the -intercepts for the circle given by the equation:

$-2\ and\ 4$

$0\ and\ 9$

$0\ and\ 1$

$-4\ and\ 2$

$1\ and\ 9$

Explanation

To find the -intercepts (where the graph crosses the -axis), we must set . This gives us the equation:

Because the left side of the equation is squared, it will always give us a positive answer. Thus if we want to take the root of both sides, we must account for this by setting up two scenarios, one where the value inside of the parentheses is positive and one where it is negative. This gives us the equations:

$x-1=\sqrt{9}$ and $x-1=-\sqrt{9}$

We can then solve these two equations to obtain $x=-2\ or\ 4$ .

Find the -intercepts for the circle given by the equation:

$-2\ and\ 4$

$0\ and\ 9$

$0\ and\ 1$

$-4\ and\ 2$

$1\ and\ 9$

Explanation

To find the -intercepts (where the graph crosses the -axis), we must set . This gives us the equation:

$x-1=\sqrt{9}$ and $x-1=-\sqrt{9}$

We can then solve these two equations to obtain $x=-2\ or\ 4$ .

Find the radius of the circle given by the equation:

Explanation

To find the center or the radius of a circle, first put the equation in the standard form for a circle: $(x-x_{1})^2+(y-y_{1})^2=r^2$ , where is the radius and $(x_{1},y_{1})$ is the center.

From our equation, we see that it has not yet been factored, so we must do that now. We can use the formula .

, so .

and , so and .

Therefore, .

Because the constant, in this case 4, was not in the original equation, we need to add it to both sides:

Now we do the same for :

We can now find :

Find the radius of the circle given by the equation:

Explanation

To find the center or the radius of a circle, first put the equation in the standard form for a circle: $(x-x_{1})^2+(y-y_{1})^2=r^2$ , where is the radius and $(x_{1},y_{1})$ is the center.

From our equation, we see that it has not yet been factored, so we must do that now. We can use the formula .

, so .

and , so and .

Therefore, .

Because the constant, in this case 4, was not in the original equation, we need to add it to both sides:

Now we do the same for :

We can now find :

Find the center of the circle given by the equation:

$\left ( 3,-9 \right )$

$\left ( 6,-18 \right )$

$\left ( 12,3 \right )$

$\left ( -8,5 \right )$

$\left ( 5,5 \right )$

Explanation

To find the center or the radius of a circle, first put the equation in standard form: $(x-x_{1})^2+(y-y_{1})^2=r^2$ , where is the radius and $(x_{1},y_{1})$ is the center.

From our equation, we see that it has not yet been factored, so we must do that now. We can use the formula .

, so .

and , so and .

This gives .

Because the constant, in this case 9, was not in the original equation, we must add it to both sides:

Now we do the same for :

We can now find the center: (3, -9)

Find the center of the circle given by the equation:

$\left ( 3,-9 \right )$

$\left ( 6,-18 \right )$

$\left ( 12,3 \right )$

$\left ( -8,5 \right )$

$\left ( 5,5 \right )$

Explanation

To find the center or the radius of a circle, first put the equation in standard form: $(x-x_{1})^2+(y-y_{1})^2=r^2$ , where is the radius and $(x_{1},y_{1})$ is the center.

From our equation, we see that it has not yet been factored, so we must do that now. We can use the formula .

, so .

and , so and .

This gives .

Because the constant, in this case 9, was not in the original equation, we must add it to both sides:

Now we do the same for :

We can now find the center: (3, -9)

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