Problem 38 - Algebra 2

Question

Write the equation of the circle that passes through the given point and has a center at the origin. (Hint: You can use the distance formula to find the radius.) $$ (\sqrt{3}, 0) $$

Step-by-Step Solution

Answer

The equation of the circle that passes through the point (\( \sqrt{3} \), 0) and has its center at the origin is \(x^2 + y^2 = 3\).

Step 1: Recall the general equation of a circle centered at origin.

The general equation of a circle centered at the origin (0,0) is of the form \(x^2 + y^2 = r^2\), where r represents the radius of the circle.

Step 2: Calculate the radius using the distance formula.

The distance formula is given by \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\). In this case, our origin (0, 0) represent the coordinates \(x_1\) and \(y_1\) and the given point (\( \sqrt{3} \), 0) represent \(x_2\) and \(y_2\). Substituting these values, we have \(d = \sqrt{(\sqrt{3} - 0)^2 + (0 - 0)^2} = \sqrt{3}\). This is our radius.

Step 3: Substitute the radius into the circle equation.

The radius \(r\) is equal to \( \sqrt{3} \). Substitute this into the equation of the circle we obtained in step 1. This gives us \(x^2 + y^2 = (\sqrt{3})^2\), which simplifies to \(x^2 + y^2 = 3\). This is the equation of our circle.

Key Concepts

Center at the Origin

When working with circles on a coordinate plane, the "center at the origin" concept is crucial. The origin refers to the point \((0, 0)\) in the coordinate system. A circle centered at the origin has a certain elegance due to its symmetrical properties.
The equation of a circle with its center at the origin is simplified because the coordinates of the center \((h, k)\) are both zeros. This means we can use the standard circle equation in its simplest form, i.e., \(x^2 + y^2 = r^2\), where \(r\) is the radius. Setting your circle's center at the origin helps reduce complexity when calculating other properties like radius and area.

Distance Formula

The distance formula is a powerful tool in geometry. It helps determine the distance between two points on a coordinate plane. When you know the coordinates of these points, you can easily compute the distance separating them.
The formula itself is derived from the Pythagorean theorem and can be expressed as:

• \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)

This equation finds its application in the circle problem when calculating the radius. Since the circle's center is at the origin \((0, 0)\), finding the radius is as simple as applying the distance formula between \((0, 0)\) and a given point like \((\sqrt{3}, 0)\). The result gives you the radius needed to write the circle's equation.

Radius

The radius is a fundamental element in the study of circles. It is the distance from the center of the circle to any point on its circumference. In our specific exercise, we found the radius using the distance formula.
With the center at \((0, 0)\) and the circle passing through the point \((\sqrt{3}, 0)\), the distance between these points is simply \(\sqrt{3}\). This value then represents the radius of the circle. A clear understanding of the radius helps in forming the mathematical relationship present in the circle equation.

General Equation of a Circle

The general equation of a circle is a powerful way to represent a circle algebraically. When the circle is centered at the origin, this equation takes the form:

• \( x^2 + y^2 = r^2 \)

Here, \(r\) is the radius of the circle. In the example provided, once we calculate \(r = \sqrt{3}\), we substitute this value into the equation.This gives us:

• \( x^2 + y^2 = (\sqrt{3})^2 \)
• Simplifying further, we arrive at \( x^2 + y^2 = 3 \)

Thus, we've effectively used the center, radius, and distance formula to derive the circle's equation, making the process intuitive and grounded in mathematical theory.