Problem 70 - Algebra and Trigonometry

Question

\(65-72\) . Show that the equation represents a circle, and find the center and radius of the circle. $$ x^{2}+y^{2}+2 x+y+1=0 $$

Step-by-Step Solution

Answer

The circle's center is \((-1, -\frac{1}{2})\) and the radius is \(\frac{1}{2}\).

Step 1: Identify the Equation Form

The given equation is \( x^2 + y^2 + 2x + y + 1 = 0 \). It resembles the general form of a circle equation, which is \( x^2 + y^2 + Dx + Ey + F = 0 \).

Step 2: Complete the Square on x

For the term involving \(x\), split \(2x\) into \((x + 1)^2 - 1\). Thus, the \(x\)-terms transform from \(x^2 + 2x\) to \((x + 1)^2 - 1\).

Step 3: Complete the Square on y

For the term involving \(y\), split \(y\) into \((y + \frac{1}{2})^2 - \frac{1}{4}\). Thus, the \(y\)-terms transform from \(y^2 + y\) to \((y + \frac{1}{2})^2 - \frac{1}{4}\).

Step 4: Rewriting the Equation

Substitute the completed square forms back into the equation: \((x + 1)^2 - 1 + (y + \frac{1}{2})^2 - \frac{1}{4} + 1 = 0\). Simplify this to \((x + 1)^2 + (y + \frac{1}{2})^2 = \frac{1}{4}\).

Step 5: Identify the Circle's Center and Radius

The standard form of the circle is \((x - h)^2 + (y - k)^2 = r^2\). Comparing with our equation \((x + 1)^2 + (y + \frac{1}{2})^2 = \frac{1}{4}\), the center \((h, k)\) is \((-1, -\frac{1}{2})\) and the radius \(r\) is \(\frac{1}{2}\).

Key Concepts

Completing the Square

When working with equations involving squares, converting them into a more readable and understandable format is often helpful. This is where the technique called "Completing the Square" comes in handy. It allows us to transform a quadratic equation with terms like

• \(x^2 + bx\)
• \(y^2 + dy\)

into perfect square trinomials. This helps in revealing the nature of the graph it represents, such as a circle.
To complete the square:
1. You group the terms of the variable.
2. For a term like \(x^2 + bx\), take half of the coefficient of \(x\), square it, and add and subtract it inside the equation. For example, with the term \(2x\), half of \(2\) is \(1\) and \(1^2 = 1\), leading to the transformation \((x + 1)^2 - 1\).
3. Do a similar process for \(y^2 + dy\). For \(y\), we add and subtract \((\frac{1}{2})^2 = \frac{1}{4}\) yielding \((y + \frac{1}{2})^2 - \frac{1}{4}\).
This method reveals the standard form of a circle equation, allowing us to easily find the center and the radius.

Center of a Circle

The center of a circle is a key characteristic that defines its position in a coordinate plane. For a circle's equation in the standard form \[(x - h)^2 + (y - k)^2 = r^2,\]- The center is represented by the coordinates \((h, k)\).
Finding the center involves comparing the transformed equation to the standard form. In our example, after completing the square, the equation becomes
\[(x + 1)^2 + (y + \frac{1}{2})^2 = \frac{1}{4}.\]
This can be rewritten as
\[(x - (-1))^2 + (y - (-\frac{1}{2}))^2 = \frac{1}{4}.\]
Thus, the center \((h, k)\) is at \((-1, -\frac{1}{2})\). Knowing the center is helpful in sketching a circle accurately on a graph.

Radius of a Circle

The radius of a circle is the distance from its center to any point on the circle's boundary. In a circle's standard form equation \[(x - h)^2 + (y - k)^2 = r^2,\]- The radius is determined by \(r\).
It is important to note that this term is squared in the equation. This means, to find the actual radius, you take the square root of whatever is on the right side of the equation.
From the example equation \[(x + 1)^2 + (y + \frac{1}{2})^2 = \frac{1}{4},\]
we can determine that \(r^2 = \frac{1}{4}\). Thus, the radius \(r\) is the square root of \(\frac{1}{4}\), which is \(\frac{1}{2}\).
The radius gives a sense of the size of the circle and is essential for plotting or interpreting the graph.