Problem 23 - Beginning and Intermediate Algebra

Question

Find an equation of the circle with the given center and radius. Center \((0,0) ;\) radius \(=\sqrt{10}\)

Step-by-Step Solution

Answer

The equation of the circle with center \((0,0)\) and radius \(\sqrt{10}\) is \(x^2 + y^2 = 10\).

Step 1: Write the general equation of a circle

To find the equation of a circle, we know that the general equation is: \[(x - h)^2 + (y - k)^2 = r^2\] where \((h, k)\) is the center of the circle and \(r\) is the radius.

Step 2: Substitute the given center and radius

In our case, the center \((h, k) = (0, 0)\) and the radius \(r = \sqrt{10}\). To find the equation of the circle, substitute these values in the general equation: \[(x - 0)^2 + (y - 0)^2 = (\sqrt{10})^2\]

Step 3: Simplify the equation

Simplify the equation: \[x^2 + y^2 = 10\]

Step 4: Write down the final equation of the circle

The equation of the circle with center \((0,0)\) and radius \(\sqrt{10}\) is: \[x^2 + y^2 = 10\]

Key Concepts

General Equation of a Circle

Understanding the general equation of a circle is crucial when determining the equation of any circle. In this formula, the equation is expressed as:

• \[(x - h)^2 + (y - k)^2 = r^2\]

The variables \(h\) and \(k\) represent the coordinates of the center of the circle, meaning where the center of the circle lies on the coordinate plane.
The \(r\) in the equation stands for the radius. The radius is the distance from the center of the circle to any point on its boundary.
This equation helps in depicting the perfect circular shape on a coordinate plane by ensuring that every point on the circle's edge is equidistant from its center.

Circle Center and Radius

To solve problems involving circles, knowing how to identify and use the center and radius is essential. When given a center \(h, k\), it represents the point from which all points on the circle are equidistant. If a circle’s center is at \(0,0\), it is centered at the origin, which greatly simplifies calculations.
The radius, labeled \(r\), is simply the circle's constant distance from the center to its boundary. In our example, the given radius is \sqrt{10}\, which means the circle’s boundary points lie \sqrt{10}\ units away from the center. By squaring the radius, which converts \(\sqrt{10}\) to 10, we use this precise measurement in the equation, ensuring the circle maintains its round shape and size.

Substitute Values in Equations

The practice of substituting values in equations allows us to personalize general formulas to match the specifics of a problem. In the circle equation, you substitute the center values \(h, k\) and the radius \(r\) into the general equation format

• \[(x - h)^2 + (y - k)^2 = r^2\]

By substituting \(h = 0\) and \(k = 0\), and \(r = \sqrt{10}\), into the equation, we simplify it to:

• \[(x - 0)^2 + (y - 0)^2 = (\sqrt{10})^2\]

This simplifies further to \(x^2 + y^2 = 10\), yielding a precise equation for the given circle. This skill of substituting values makes the process straightforward and applicable to any circle-related problem, essentially turning the general into the specific.