Circles: Videos & Practice Problems

Circles are defined as the set of all points equidistant from a center point, characterized by the radius. The standard equation for a circle centered at the origin is x^2 + y^2 = r^2. For circles not at the origin, the equation is x - h + y - k = r^2, where (h, k) is the center. Converting from general to standard form involves completing the square for both x and y terms.

Circles in Standard Form

Conic sections are fascinating shapes formed by slicing a three-dimensional cone with a two-dimensional plane, and one of the most fundamental shapes is the circle. A circle is defined as the set of all points that are equidistant from a central point, known as the center. This unique property of circles allows us to describe their size and position using specific mathematical equations.

To graph a circle, two key components are required: the center and the radius. The center is typically denoted as (h, k), where h represents the horizontal position and k represents the vertical position. For example, if a circle is centered at the origin (0, 0), the equation of the circle can be expressed as:

x^2 + y^2 = r^2

Here, r is the radius of the circle. If the circle is not centered at the origin, the equation adjusts to account for the new center, resulting in the form:

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

For instance, if a circle has a center at (2, 1) and a radius of 4, the equation would be:

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

When given the equation of a circle, one can derive the center and radius to graph it. The center can be identified directly from the equation, while the radius can be determined by taking the square root of the constant on the right side of the equation. For example, if the equation is:

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

the center is (1, 2) and the radius is 3, since \( r^2 = 9 \) implies \( r = \sqrt{9} = 3 \).

To graph the circle, one would plot points at a distance of the radius from the center in all four cardinal directions (up, down, left, right) and then connect these points with a smooth curve to form the circle.

It is important to note that a circle does not represent a function in the mathematical sense, as it fails the vertical line test. This test states that if a vertical line intersects a graph at more than one point, the graph does not represent a function. Since a circle can be intersected by a vertical line at two points, it is classified as a non-function.

Understanding the properties and equations of circles is essential in geometry, as it lays the groundwork for exploring more complex conic sections and their applications in various fields.

Sketch a graph of the circle based on the following equation: $x^2+\left(y-1\right)^2=9$

• A.
• B.
• C.
• D.

Answer: (D)

Sketch a graph of the circle based on the following equation: $\left(x-2\right)^2+\left(y+3\right)^2=1$

• A.
• B.
• C.
• D.

Answer: (B)

Find the equation for the following circle:

• A. $x^2+y^2=4$
• B. $\left(x+1\right)^2+y^2=4$
• C. $\left(x-1\right)^2+y^2=4$
• D. $\left(x+1\right)^2+y^2=16$

Answer: (D)

Circles in Standard Form Example 1

To write the equation of a circle, we utilize the standard form given by the equation:

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

In this equation, \((h, k)\) represents the center of the circle, and \(r\) is the radius. Let's explore how to apply this formula through a few examples.

In the first scenario, the center of the circle is at \((0, 1)\) and the radius is \(3\). Substituting these values into the standard form, we have:

\( (x - 0)^2 + (y - 1)^2 = 3^2 \)

This simplifies to:

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

Next, consider a circle with its center at \((1, -2)\) and a radius of \(1\). Plugging these values into the standard form gives us:

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

This results in the equation:

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

For the final example, the center is at \((-6, 3)\) and the radius is \(\sqrt{5}\). The equation becomes:

\( (x + 6)^2 + (y - 3)^2 = (\sqrt{5})^2 \)

Which simplifies to:

\( (x + 6)^2 + (y - 3)^2 = 5 \)

By understanding the relationship between the center, radius, and the standard form of the circle's equation, one can easily derive the equations and sketch the corresponding graphs. This method is essential for visualizing and solving problems related to circles in coordinate geometry.

Circles in General Form

In the study of circles, understanding the general form of a circle's equation is crucial for converting it to standard form. The general form of a circle's equation can appear daunting, but with the right approach, it can be simplified. The general form is typically expressed as:

$$x^2 + y^2 + Dx + Ey + F = 0$$

To convert this equation into standard form, which is given by:

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

where \((h, k)\) is the center of the circle and \(r\) is the radius, you will need to complete the square for both the \(x\) and \(y\) terms. The process begins by grouping the \(x\) and \(y\) terms together and moving the constant term to the other side of the equation. For example, if you have:

$$x^2 + 2x + y^2 - 4y + 1 = 0$$

you would rearrange it to:

$$x^2 + 2x + y^2 - 4y = -1$$

Next, you complete the square for the \(x\) terms. Take the coefficient of \(x\) (which is 2), divide it by 2 to get 1, and then square it to obtain 1. You add this value to both sides of the equation. Repeat this for the \(y\) terms, where the coefficient is -4. Dividing by 2 gives -2, and squaring it results in 4. Add this to both sides as well:

$$x^2 + 2x + 1 + y^2 - 4y + 4 = -1 + 1 + 4$$

This simplifies to:

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

Now, you can identify the center of the circle as \((-1, 2)\) and the radius \(r\) as the square root of 4, which is 2. This means the circle can be graphed by starting at the center and moving 2 units in all directions (up, down, left, right) to find the points on the circle. Connecting these points with a smooth curve will yield the graph of the circle.

In summary, converting from general to standard form involves grouping terms, completing the square, and identifying the center and radius for graphing. Mastering this process is essential for effectively working with circles in geometry.

Determine if the equation $x^2+y^2-2x+4y-4=0$ is a circle, and if it is, find its center and radius.

• A. Is a circle, center = $c\left(0,0\right)$, radius $r=2$.
• B. Is a circle, center = $c\left(0,0\right)$, radius $r=3$.
• C. Is a circle, center = $c\left(1,-2\right)$, radius $r=3$ .
• D. Is not a circle.

Answer: (C)

Determine if the equation $x^3+y^2+4x-8y+4=0$ is a circle, and if it is, find its center and radius.

• A. Is a circle, center = $c\left(0,0\right)$, radius $r=4$.
• B. Is a circle, center = $c\left(0,0\right)c\left(2,-4\right)$, radius $r=4$.
• C. Is a circle, center = $c\left(-2,4\right)c\left(0,0\right)$, radius $r=4$.
• D. Is not a circle.

Answer: (D)

Here’s what students ask on this topic:

What is the standard equation of a circle?

The standard equation of a circle is given by:

$x^2+y^2=r^2$

where $r$ is the radius of the circle and the center is at the origin (0, 0). If the circle is not centered at the origin, the equation becomes:

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

where $h$ and $k$ are the coordinates of the center of the circle.

How do you find the center and radius of a circle from its equation?

To find the center and radius of a circle from its equation, you need to identify the standard form of the circle's equation:

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

The center of the circle is given by the coordinates $(h,k)$, and the radius is the square root of the constant term on the right side of the equation, $\sqrt{r}$.

What is the general form of a circle's equation?

The general form of a circle's equation is:

$x^2+y^2+\mathrm{Dx}+\mathrm{Ey}+F=\; 0$

where $D$, $E$, and $F$ are constants. To convert this to the standard form, you need to complete the square for both $x$ and $y$ terms.

How do you convert the general form of a circle's equation to the standard form?

To convert the general form $x^2+y^2+\mathrm{Dx}+\mathrm{Ey}+F=\; 0$ to the standard form, follow these steps:

1. Group the $x$ and $y$ terms together.
2. Move the constant term $F$ to the other side of the equation.
3. Complete the square for both $x$ and $y$ terms.
4. Rewrite the equation in the form $(x-h)2^2+\; (y-k)2^2=r^2$.

How do you graph a circle given its equation in standard form?

To graph a circle given its equation in standard form $(x-h)2^2+\; (y-k)2^2=r^2$, follow these steps:

1. Identify the center $(h,k)$ and the radius $\sqrt{r}$.
2. Plot the center on the coordinate plane.
3. From the center, move $r$ units up, down, left, and right to mark four points.
4. Draw a smooth curve connecting these points to form the circle.