Problem 42 - Precalculus

Question

Specify the center and radius of each circle. Also, determine whether the given point lies on the circle. $$x^{2}+y^{2}=1 ;(1 / 2, \sqrt{3} / 2)$$

Step-by-Step Solution

Answer

The circle has center (0, 0) and radius 1; the point lies on the circle.

Step 1: Identify the Standard Form of a Circle's Equation

The standard form of a circle's equation is \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius. Our given equation is \(x^2 + y^2 = 1\).

Step 2: Find the Center and Radius of the Circle

Comparing \(x^2 + y^2 = 1\) with the standard form \((x-h)^2 + (y-k)^2 = r^2\), we identify that \(h = 0\), \(k = 0\), and \(r^2 = 1\). Thus, the center is \((0, 0)\) and the radius \(r = \sqrt{1} = 1\).

Step 3: Verify if the Point Lies on the Circle

A point \((x, y)\) lies on the circle if it satisfies the equation \(x^2 + y^2 = 1\). Substitute \(x = \frac{1}{2}\) and \(y = \frac{\sqrt{3}}{2}\) into the equation: \(\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2\).

Step 4: Calculate the Verification

Compute \(\left(\frac{1}{2}\right)^2 = \frac{1}{4}\) and \(\left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}\). Adding them, we get \(\frac{1}{4} + \frac{3}{4} = 1\).

Step 5: Conclusion on the Point's Position

Since \(\frac{1}{4} + \frac{3}{4} = 1\) matches the right-hand side of the equation, the point \(\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)\) lies on the circle.

Key Concepts

Standard Form of a Circle

The concept of the standard form of a circle's equation is key to understanding circles in geometry. It is expressed as \((x - h)^2 + (y - k)^2 = r^2\). Here, \((h, k)\) represents the center of the circle, and \(r\) denotes its radius. The standard form centers the circle at specific coordinates and makes it easier to analyze its properties.

When examining the equation \(x^2 + y^2 = 1\), we can see this is a simplified version of the standard form where \(h\) and \(k\) are equal to zero. This tells us that the center of the circle is at the origin. Simplification typically occurs when the circle's center appears on coordinate axes. Recognizing these shortcuts can make solving problems faster and more intuitive.

Center of a Circle

The center of a circle is a crucial element that determines its location in a coordinate plane. It is defined by the coordinates \((h, k)\) in the standard form equation. For the equation \(x^2 + y^2 = 1\), the center is easily identified by looking at the terms \((x - h)^2\) and \((y - k)^2\).

In this specific case, there are no \(h\) or \(k\) subtracted from \(x\) or \(y\), implying both values are zero. Thus, we conclude:

• The center of the circle is at the point \((0, 0)\).

This origin-centered circle is symmetric around the axes, making calculations involving points on the circle straightforward.

Radius of a Circle

The radius of a circle measures the distance from its center to any point on the circle. It is represented by \(r\) in the equation \((x - h)^2 + (y - k)^2 = r^2\). In our given circle equation \(x^2 + y^2 = 1\), by comparison, \(r^2 = 1\), meaning the radius \(r\) is \[r = \sqrt{1} = 1\].

This tells us every point on the circle is exactly one unit away from the center \((0, 0)\). Understanding the radius helps us ascertain volume, circumference, and other important circle properties efficiently.

Point on a Circle

To determine if a given point lies on the circle, one must check if this point satisfies the circle's equation. Consider the point \((\frac{1}{2}, \frac{\sqrt{3}}{2})\) for the circle defined by \(x^2 + y^2 = 1\). Substituting \(x = \frac{1}{2}\) and \(y = \frac{\sqrt{3}}{2}\) into the equation allows verification:

• Calculate \(\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2\).
• It becomes \(\frac{1}{4} + \frac{3}{4} = 1\).

The result, \(1\), matches the right hand side of the circle's equation, confirming:

• The point \(\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)\) is indeed on the circle.

This simple check assures us of the point’s position relative to the circle, utilizing basic substitutions and algebraic logic.