Problem 62 - Calculus for Biology and Medicine

Question

Find the center and the radius of the circle given by the equation $$ x^{2}+y^{2}+2 x-4 y+1=0 $$

Step-by-Step Solution

Answer

Center: \((-1, 2)\); Radius: 2.

Step 1: Identify the general form of a circle's equation

The general 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. We need to reformat the given equation into this form.

Step 2: Complete the square for the x-terms

Start by rearranging the x-terms: \(x^2 + 2x\). To complete the square, take half of 2, square it to get 1, and add/subtract 1 inside the equation: \(x^2 + 2x = (x + 1)^2 - 1\).

Step 3: Complete the square for the y-terms

Now, rearrange the y-terms: \(y^2 - 4y\). Take half of -4, square it to get 4, and add/subtract 4: \(y^2 - 4y = (y - 2)^2 - 4\).

Step 4: Substitute completed squares back into the equation

Substitute the completed squares into the original equation: \[(x + 1)^2 - 1 + (y - 2)^2 - 4 + 1 = 0\]Simplify it to:\[(x + 1)^2 + (y - 2)^2 = 4\]

Step 5: Identify the center and the radius

The reformed equation \((x + 1)^2 + (y - 2)^2 = 4\) is now in the standard form, meaning the center \((h, k)\) is \((-1, 2)\) and the radius \(r\) is \(\sqrt{4} = 2\).

Key Concepts

Completing the Square

Completing the square is an algebraic technique used to transform quadratic expressions into perfect square trinomials. It simplifies the equation into a format that reveals more about the graph of the function. In the context of circle equations, this method helps rearrange the equation into a recognizable form.
For a quadratic expression such as \(x^2 + 2x\), the steps to complete the square are as follows:

• Take half of the linear coefficient (in this case, 2), which results in 1.
• Square this result: \(1^2 = 1\).
• Add and subtract this square in the equation to maintain equality; thus the expression becomes: \((x + 1)^2 - 1\).

This adjustment makes it clear that the expression is a perfect square trinomial minus an adjustment factor. Similarly, for the y-term, \(y^2 - 4y\), the process involves taking half of \(-4\), squaring to get 4, and adjusting the equation as \((y - 2)^2 - 4\). These steps are crucial for converting an equation into its standard circle form.

Standard Form of a Circle

The standard form of a circle's equation is a neat package that shows the circle's key features clearly. It's written as \((x - h)^2 + (y - k)^2 = r^2\). Here's what each component means:

• \((h, k)\) is the center of the circle.
• \(r\) is the radius of the circle, which is always a non-negative number.

When you start with a complex or jumbled equation, the goal is to use methods like completing the square to rewrite it in standard form. For instance, if you start with \(x^2 + y^2 + 2x - 4y + 1 = 0\), completing the square helps reshape it into \((x + 1)^2 + (y - 2)^2 = 4\). Now, it’s clear and reveals the circle's geometric features without further digging. This form is convenient and easy to interpret for anyone looking to understand how the circle behaves in the coordinate plane.

Center and Radius

Identifying the center and radius from a circle equation in standard form is straightforward. Simply look at the terms within the parentheses and the radius squared term.
The equation \((x + 1)^2 + (y - 2)^2 = 4\) is already in standard form. From here, you can extract:

• The center of the circle, \((h, k)\), is found by reversing the signs inside the parentheses: \((-1, 2)\).
• The term on the right side of the equation, the number 4, represents \(r^2\). Taking the square root gives the radius \(r = 2\).

This extraction process makes understanding the geometry of the circle intuitive and allows one to visually place the circle correctly in the coordinate plane. Remember that the center points \((h, k)\) and the radius \(r\) define the shape and position of the circle completely, making this form highly valuable for graphing and analysis.