Problem 8
Question
Find the standard form of the equation of the circle with the given characteristics. Center at origin; radius: \(\sqrt{11}\)
Step-by-Step Solution
Verified Answer
The standard form of the equation of the circle with given characteristics is \(x^2 + y^2 = 11\).
1Step 1: Identify the given parameters
The center of the circle is at the origin, and the radius of the circle is \(\sqrt{11}\). So, we have the center coordinates \((0,0)\) and radius \(r = \sqrt{11}\).
2Step 2: Apply the values in the circle's equation
The equation of a circle with center at the origin in two-dimensional space is given by \(x^2 + y^2 = r^2\). Substituting our given radius \(\sqrt{11}\) into the equation, we get \(x^2 + y^2 = (\sqrt{11})^2\).
3Step 3: Simplify the equation
On simplifying, \((\sqrt{11})^2\) gives us 11. Therefore, the equation of the circle in standard form is \(x^2 + y^2 = 11\).
Key Concepts
Center at OriginRadiusStandard Form of a Circle
Center at Origin
Whenever you learn about equations of circles, the location of the circle's center is crucial. When we talk about a circle with its **center at the origin**, we're referring to a specific point on a coordinate plane. The origin is the point
Any other points on the circle are at a consistent distance, known as the radius, from this center point.
The simplicity of the origin as a center means the equation doesn't need to account for shifts along the x or y-axis. This is why the circle's formula becomes more straightforward, without additional terms to compensate for location shifts.
- where the x-axis and y-axis intersect,
- and it is represented by the coordinates \((0, 0)\).
Any other points on the circle are at a consistent distance, known as the radius, from this center point.
The simplicity of the origin as a center means the equation doesn't need to account for shifts along the x or y-axis. This is why the circle's formula becomes more straightforward, without additional terms to compensate for location shifts.
Radius
Understanding the **radius** is key to diving deeper into the mathematics of a circle. The radius is the distance from the center of the circle to any point on its boundary.
In this exercise, the radius is given as \(\sqrt{11}\). This tells us that:
The squared radius,\((\sqrt{11})^2\), uses simple arithmetic to turn the radial distance into \(11\).
In essence, it acts as a scalar factor in the equation, shaping the size of the circle on the coordinate plane.
In this exercise, the radius is given as \(\sqrt{11}\). This tells us that:
- Every point on the circle's outline is exactly \(\sqrt{11}\) units away from the origin.
The squared radius,\((\sqrt{11})^2\), uses simple arithmetic to turn the radial distance into \(11\).
In essence, it acts as a scalar factor in the equation, shaping the size of the circle on the coordinate plane.
Standard Form of a Circle
One of the most common ways to represent a circle in coordinates is by using its **standard form** equation.
This standard form helps to easily identify the center and radius of the circle. For a circle centered at the origin, the equation simplifies to:
For this specific exercise, since the center is at \((0, 0)\) and \(r = \sqrt{11}\),
we substitute \(r^2\) with \(11\):
This standard form helps to easily identify the center and radius of the circle. For a circle centered at the origin, the equation simplifies to:
- \[x^2 + y^2 = r^2\],
For this specific exercise, since the center is at \((0, 0)\) and \(r = \sqrt{11}\),
we substitute \(r^2\) with \(11\):
- The equation becomes\[x^2 + y^2 = 11.\]
Other exercises in this chapter
Problem 7
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Consider the ellipse given by \(\frac{x^{2}}{2^{2}}+\frac{y^{2}}{8^{2}}=1.\) Is the ellipse elongated or nearly circular?
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