Problem 69
Question
65–72 ? Show that the equation represents a circle, and find the center and radius of the circle. $$ x^{2}+y^{2}+x=0 $$
Step-by-Step Solution
Verified Answer
The circle has center \((-\frac{1}{2}, 0)\) and radius \(\frac{1}{2}\).
1Step 1: Rearrange the Equation
The given equation is \( x^2 + y^2 + x = 0 \). To identify it as a circle, we rearrange it to focus on the terms involving \( x \) and \( y \). Separate the variables in the equation: \( x^2 + x + y^2 = 0 \).
2Step 2: Complete the Square for x
Identify the terms involving \( x \), which are \( x^2 + x \). To complete the square, add and subtract \( \left( \frac{1}{2} \right)^2 = \frac{1}{4} \): \( x^2 + x + \frac{1}{4} - \frac{1}{4} + y^2 = 0 \). This can be rewritten as \( (x + \frac{1}{2})^2 - \frac{1}{4} + y^2 = 0 \).
3Step 3: Simplify the Equation
Move the \( -\frac{1}{4} \) to the other side of the equation: \( (x + \frac{1}{2})^2 + y^2 = \frac{1}{4} \).
4Step 4: Identify the Center and Radius of the Circle
The equation is now in the standard form of a circle: \( (x - h)^2 + (y - k)^2 = r^2 \). Comparing, we have \( h = -\frac{1}{2} \), \( k = 0 \), and \( r = \frac{1}{2} \). Thus, the center of the circle is \( \left( -\frac{1}{2}, 0 \right) \) and the radius is \( \frac{1}{2} \).
Key Concepts
Completing the SquareCenter of a CircleRadius of a Circle
Completing the Square
Completing the square is a crucial algebraic technique used to transform a quadratic expression into a perfect square trinomial. This method is particularly useful when working with equations of circles, as it helps to identify the circle's center and radius.
To complete the square for a term like \( x^2 + x \), follow these steps:
To complete the square for a term like \( x^2 + x \), follow these steps:
- Identify the coefficient of the \( x \) term, which is 1 in this case.
- Divide the coefficient by 2, giving you \( \frac{1}{2} \).
- Square this result to get \( \left( \frac{1}{2} \right)^2 = \frac{1}{4} \).
- Add and subtract \( \frac{1}{4} \) within the equation to balance it, forming a complete square as \( (x + \frac{1}{2})^2 \).
Center of a Circle
In the standard equation of a circle, \( (x - h)^2 + (y - k)^2 = r^2 \), the terms \( h \) and \( k \) represent the center's coordinates \((h, k)\) of the circle. Knowing the center is essential to understanding the position and orientation of a circle in the coordinate plane.
For our specific example, after completing the square, we rewrite the equation as \( (x + \frac{1}{2})^2 + (y - 0)^2 = \frac{1}{4} \).
For our specific example, after completing the square, we rewrite the equation as \( (x + \frac{1}{2})^2 + (y - 0)^2 = \frac{1}{4} \).
- The expression \( (x + \frac{1}{2})^2 \) indicates that \( h = -\frac{1}{2} \).
- The \( y^2 \) term simplifies directly to \( (y - 0)^2 \), showing that \( k = 0 \).
Radius of a Circle
The radius is a fundamental characteristic of a circle. It represents the distance from the center of the circle to any point on its circumference. In the circle's equation form \( (x - h)^2 + (y - k)^2 = r^2 \), the radius is denoted by \( r \).
From our simplified equation \( (x + \frac{1}{2})^2 + y^2 = \frac{1}{4} \), it's clear that:
From our simplified equation \( (x + \frac{1}{2})^2 + y^2 = \frac{1}{4} \), it's clear that:
- The right side of the equation is \( \frac{1}{4} \).
- This value corresponds to \( r^2 \), so the radius \( r \) is \( \sqrt{\frac{1}{4}} \).
- Calculating further gives \( r = \frac{1}{2} \).
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