Circles - Intermediate Geometry

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Question

A circle has a diameter starting at and ending at . What is the equation of the circle?

Answer

Using the distance formula, the diameter is 10 units long, so the radius is 5. Then, by finding the midpoint of the diameter, you know the center of the cirle. Plugging values into the circle equation yields the final answer.

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Question

What is the equation for a circle centered at with a radius of ?

Answer

The general equation for a circle centered at the origin is given by $x^{2}+y^{2}=r^{2}$ where is the radius of the circle.

To translate the origin to the first quadrant we need to subtract the appropriate amount to bring it back to center.

So the equation becomes $(x-1)^{2}+(y-2)^{2}=9$

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Question

What is the equation of a circle with diameter , that is tangent to the x-axis at the point and the y-axis at the point ?

Answer

The equation for a circle is where the center of the circle lies at the point and the radius of the circle is . If we are looking for a circle with a diameter of , then its radius must be . For the circle to be tangent to the x-axis at the point and the y-axis at , it must be centered at the point . Therefore, the equation of the circle will be .

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Question

A circle has its center at the point and a radius of units.

What is the equation of the circle?

Answer

The equation for a circle is

where (h, k) is the center of the circle and r is the radius.

Plugging in the values of , , and , we get

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Question

Find the equation of a circle if the radius of the circle is and the center is located at the origin.

Answer

The formula for the equation of a circle is:

The values of $\left ( h,k\right )$ represent the center, and both values are zero at the origin.

Plug in the known values and reduce.

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Question

Write the equation for a circle with the domain of and a range of .

Answer

The domain of a circle gives the width of the circle, the range of a circle gives the height of a circle - both of which equal the diameter.

The domain and range numbers are both 6 apart, meaning that the diameter of the circle is 6, which means the radius of the circle is 3.

The halfway point between the domain gives the x-coordinate for the center of the circle, which in this case is 5. 5 is halfway between 2 and 8.

The halfway point between the range gives the y-coordinate for the center of the circle, which in this case is 2. 2 is halfway between -1 and 5.

Now we have all of the information needed to plug into the equation for a circle in standard form as shown below where r is the radius and the center is (h,k):

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

Given a circle with radius of 3, and center of (5,2) we get the below equation of a circle.

$(x-5)^{2}+(y-2)^{2}=9$

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Question

If the point (2,4) is on a circle whose center is at (6,8), which of the following is the standard form of the equation of this circle?

Answer

The standard equation of a circle is where is the center and r is the radius. Since we were given a point on the circumference of the circle and its center, we use the distance formula to find the radius and then plug the radius and our center point into our equation.

$r=\sqrt{(6-2)^2+(8-4)^2}=\sqrt{32}$

Now substitute the center points and radius into the standard equation of a circle:

$(x-6)^2+(y-8)^2=(\sqrt{32})^2\rightarrow{(x-6)^2+(y-8)^2=32}$

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Question

Write the equation for a circle with center passing through the point

Answer

To determine the equation for a circle, we need to know the center and the radius. In this case, we know the center and one of the points on the circle. The radius of the circle is the distance from the center to this point, so to determine it, use the distance formula:

$r = \sqrt{(2+1)^2 + (5-3)^2} = \sqrt{9+4} = \sqrt{13}$

The equation for a circle is written as where the center is and the radius is r. For this circle, plug in and $r = \sqrt{13}$ :

$(x-2)^2 + (y-5)^2 = \sqrt{13}^2$ or more simply

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Question

Write the equation for a circle passing through the point and centered at the origin.

Answer

To determine the equation of the circle, we need to know the radius, or the distance from the origin to the point on the circle. Use the distance formula:

$r = \sqrt{(-1-0)^2 + (3-0)^2 } = \sqrt{10}$

Since the circle is centered at the origin, its equation is $(x-0)^2 + (y-0)^2 = \sqrt{10}^2$ or more simply

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Question

Which of these points is inside the circle

?

Answer

Plugging in the point shows that this point is inside the circle, since the left side of the equation will be less than the right:

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Question

Write the equation for the circle with center and radius

Answer

The equation for a circle is in the form where is the center and r is the radius.

In this case:

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Question

What is the equation for a circle centered at passing through the point

Answer

The equation of a circle is defined by where the center is and the radius is r. The radius is the distance from the center to any point on the circle, so we can use the distance formula to calculate it:

$r= \sqrt{(8+2)^2 + (0+3)^2} = \sqrt{100+9} = \sqrt{109}$

The equation is then $(x-8)^2 + (y-0)^2 = \sqrt{109}^2$ or more simplified,

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Question

Write the equation for a circle with center and passing through the point

Answer

To determine the radius of the circle, use the distance formula:

$r = \sqrt{(1-0)^2 + (-3-4)^2} = \sqrt{1+49} = \sqrt{50}$

The equation for the circle is $(x-1)^2 + (y--3)^2 = \sqrt{50}^2$ or more simply

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Question

What is the equation of a circle that has its center at and a radius of ?

Answer

Recall the standard form for the equation of a circle:

In this equation, represents the center of the circle and is the radius of the circle.

Since the center and radius of the circle are given to us, just substitute in the information itno the standard form of the equation of the circle.

The equation of the given circle is:

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Question

What is the equation of a circle that has its center at and a radius of ?

Answer

Recall the standard form for the equation of a circle:

In this equation, represents the center of the circle and is the radius of the circle.

Since the center and radius of the circle are given to us, just substitute in the information itno the standard form of the equation of the circle.

The equation of the given circle is:

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Question

What is the equation of a circle that has its center at and a radius of ?

Answer

Recall the standard form for the equation of a circle:

In this equation, represents the center of the circle and is the radius of the circle.

Since the center and radius of the circle are given to us, just substitute in the information itno the standard form of the equation of the circle.

The equation of the given circle is:

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Question

What is the equation of a circle that has its center at $(5, \frac{1}{4})$ and a radius length of $\frac{1}{5}$ ?

Answer

Recall the standard form for the equation of a circle:

In this equation, represents the center of the circle and is the radius of the circle.

Since the center and radius of the circle are given to us, just substitute in the information itno the standard form of the equation of the circle.

The equation of the given circle is:

$(x-5)^2+(y-\frac{1}{4})^2=\frac{1}{25}$

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Question

What is the equation of a circle that has its center at and a radius of $\frac{3}{5}$ ?

Answer

Recall the standard form for the equation of a circle:

In this equation, represents the center of the circle and is the radius of the circle.

Since the center and radius of the circle are given to us, just substitute in the information itno the standard form of the equation of the circle.

The equation of the given circle is:

$(x-15)^2+(y-2)^2=\frac{9}{25}$

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Question

What is the equation of a circle that has its center of and a radius of ?

Answer

Recall the standard form for the equation of a circle:

In this equation, represents the center of the circle and is the radius of the circle.

Since the center and radius of the circle are given to us, just substitute in the information itno the standard form of the equation of the circle.

The equation of the given circle is:

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Question

What is the equation of a circle that has its center of and a radius of ?

Answer

Recall the standard form for the equation of a circle:

In this equation, represents the center of the circle and is the radius of the circle.

Since the center and radius of the circle are given to us, just substitute in the information itno the standard form of the equation of the circle.

The equation of the given circle is:

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