Problem 61 - Calculus for Biology and Medicine

Question

Find the center and the radius of the circle given by the equation $$ 0=x^{2}+y^{2}-4 x+2 y-11 $$

Step-by-Step Solution

Answer

The center of the circle is (2, -1), and the radius is 4.

Step 1: Rearrange the Equation

First, rearrange the given circle equation in the standard form. Start by moving all terms involving variables to one side of the equation: \ \( x^2 - 4x + y^2 + 2y = 11 \). \ This makes it easier to complete the square in the next steps.

Step 2: Complete the Square for x

To complete the square for the \(x\) terms, take the coefficient of \(x\), which is \(-4\), divide it by 2 to get \(-2\), and then square it to obtain 4. \ Add and subtract this square value inside the equation: \ \( (x^2 - 4x + 4) + (y^2 + 2y) = 11 + 4 \). \ This simplifies to \( (x - 2)^2 + y^2 + 2y = 15 \).

Step 3: Complete the Square for y

Next, complete the square for the \(y\) terms. Take the coefficient of \(y\), which is 2, divide it by 2 to get 1, and square it to obtain 1. \ Add and subtract this square inside the equation: \ \( (x - 2)^2 + (y^2 + 2y + 1) = 15 + 1 \). \ This simplifies to \( (x - 2)^2 + (y + 1)^2 = 16 \).

Step 4: Write the Equation in Standard Form

Now, the equation is in the standard form of a circle: \( (x - h)^2 + (y - k)^2 = r^2 \). \ For this circle, \(h = 2\), \(k = -1\), and \(r^2 = 16\). \ The standard form of the equation is \( (x - 2)^2 + (y + 1)^2 = 16 \).

Step 5: Identify the Center and Radius

From the standard form equation \( (x - 2)^2 + (y + 1)^2 = 16 \), we can identify the center of the circle \((h, k)\) and the radius \(r\). \ The center is \((h, k) = (2, -1)\), and the radius is \(r = \sqrt{16} = 4\).

Key Concepts

Completing the Square

"Completing the square" is a method used to transform a quadratic expression into a perfect square trinomial. This technique is essential when working with circle equations to simplify them into a more recognizable form. Here's how it works:

• Identify the quadratic terms in the equation. Separate the terms involving the same variable together.
• For a term like \(x^2 - 4x\), take the linear coefficient (here, \(-4\)), divide it by 2, then square the result. \(-4/2 = -2\) and \((-2)^2 = 4\).
• Add and subtract this square within the equation to 'complete the square.' For example, \((x^2 - 4x + 4)\) becomes \((x - 2)^2\).
• Repeat this process for the \(y\) terms, ensuring each part of the equation is balanced by adding the square outside the perfect square trinomial as necessary.

This results in a clean expression, making it easier to identify other circle features.

Standard Form of a Circle

The "standard form of a circle" is a classic formula used to describe a circle's equation in a tidy manner. A circle in this form is expressed as:\[(x - h)^2 + (y - k)^2 = r^2\]Here, it's easy to understand:

• \((x - h)\) and \((y - k)\) represent the shifted coordinates of the circle, allowing us to identify how the circle is positioned relative to the origin.
• \(r^2\) is the square of the circle's radius, making \(r\) the distance from the circle's center to any point on its circumference.

By converting a general quadratic equation into the standard form of a circle, it becomes straightforward to understand the circle's size and location, which are represented by the center \((h, k)\) and radius \(r\). Ensuring equations fit this format aids in solving and graphing circles effectively.

Center and Radius of a Circle

Determining the "center and radius of a circle" is essential to graph and understand the circle's position and size in the coordinate plane. Once an equation is put into the standard form \[(x - h)^2 + (y - k)^2 = r^2\]you can easily identify these features:

• The center of the circle is denoted by \((h, k)\). This point is the midpoint and anchor for the circle in the plane.
• The radius is \(r\), derived from \(r^2\). Simply take the square root to find the actual radius length.

For instance, with the standard circle equation \((x - 2)^2 + (y + 1)^2 = 16\):

• The center \((h, k)\) is \((2, -1)\). This tells us that the center is 2 units right and 1 unit down from the origin.
• The radius \(r\) is \(4\), since the square root of 16 is 4, meaning the circle extends 4 units uniformly from the center.

These elements allow easy graphing and understanding of the circle's properties and location in space.