Problem 74 - Algebra and Trigonometry

Question

\(73-76\) Sketch the graph of the equation. $$ 4 x^{2}+4 y^{2}+2 x=0 $$

Step-by-Step Solution

Answer

The graph is a circle centered at \((-\frac{1}{4}, 0)\) with radius \(\frac{1}{4}\).

Step 1: Rearrange the Equation

The given equation is \(4x^2 + 4y^2 + 2x = 0\). Start by rearranging the terms to isolate \(x\). Divide every term by 4 to simplify: \(x^2 + y^2 + \frac{x}{2} = 0\).

Step 2: Complete the Square for x

In order to simplify the equation, we need to complete the square for the \(x\) terms. Take \(x^2 + \frac{x}{2}\), find \(\left(\frac{1}{4}\right)^2 = \frac{1}{16}\), and add and subtract it inside the equation to form a complete square: \(\left(x + \frac{1}{4}\right)^2 = x^2 + \frac{x}{2} + \frac{1}{16}\).

Step 3: Update the Equation

Substitute the complete square back into the equation. Thus, we have: \(\left(x + \frac{1}{4}\right)^2 - \frac{1}{16} + y^2 = 0\). Move \(-\frac{1}{16}\) to the other side to isolate the complete square form: \(\left(x + \frac{1}{4}\right)^2 + y^2 = \frac{1}{16}\).

Step 4: Recognize the Equation Type

The equation \(\left(x + \frac{1}{4}\right)^2 + y^2 = \frac{1}{16}\) is in the standard form of a circle equation: \((x - h)^2 + (y - k)^2 = r^2\), where the center is \((-\frac{1}{4}, 0)\) and the radius is \(\frac{1}{4}\).

Step 5: Sketch the Graph

Draw the circle with center at \((-\frac{1}{4}, 0)\) and a radius of \(\frac{1}{4}\). This is a small circle centered slightly to the left of the origin along the x-axis and symmetric about the origin.

Key Concepts

Completing the Square

Completing the square is a helpful method used in algebra to transform a quadratic equation into a form that is easier to work with, specifically by rewriting a trinomial of the form \(ax^2+bx+c\) into \((x-h)^2+k\). This technique is particularly valuable when dealing with circle equations, as it allows us to express them in the standard form. Let's break down the steps involved in completing the square.

• Start with a quadratic expression like \(x^2 + bx\).
• Find half of the coefficient of \(x\), which is \(b/2\), and then square it.
• Add and subtract this squared term within the expression. This creates a perfect square trinomial.

In the equation from the exercise, this method was used to transform \(x^2 + \frac{x}{2}\) into a perfect square \(\left(x + \frac{1}{4}\right)^2\). By turning it into this form, it fits neatly into the circle equation structure, allowing further progress to isolate circle characteristics such as its center and radius.

Center of a Circle

The center of a circle is a crucial concept in geometry and algebra. It is the point that is equidistant from all points on the circle. In an equation given in its standard form \((x - h)^2 + (y - k)^2 = r^2\), the coordinates \((h, k)\) indicate the center of the circle. To identify the center from an equation not in standard form, like the exercise provided, completing the square is often necessary.

• The process involves re-arranging and simplifying the equation until it resembles the standard circle equation.
• Once the equation is in the form \((x - h)^2 + (y - k)^2 = r^2\), \(h\) and \(k\) can be read directly from the terms that involve \(x\) and \(y\).

For the exercise featured, by completing the square, the center was found to be \((-\frac{1}{4}, 0)\). This indicates that the circle is positioned just slightly to the left of the y-axis, showcasing how precise transformations of equations lead to precise geometric insights.

Radius of a Circle

The radius of a circle is the distance from the center of the circle to any point on its circumference. In the standard circle equation \((x-h)^2 + (y-k)^2 = r^2\), the radius \(r\) is found by taking the square root of the value on the right-hand side of the equation. This value is crucial in defining the size of the circle. When working through algebraic transformations, the value of \(r^2\) becomes apparent once the equation reaches standard form.

• Simplify the circle equation to isolate \(r^2\) on one side.
• Take the square root of \(r^2\) to find the radius \(r\).

In the provided exercise, after completing the square and adjusting the equation to match a standard form, the radius was identified as \(\frac{1}{4}\). This small radius suggests the circle is tight and compact, located near its center, and provides a geometric dimension that is useful for graphing and spatial understanding.