Problem 20 - Technical Mathematics with Calculus

Question

Rewrite each general equation in standard form. Find the center and radius. Graph. $$x^{2}+y^{2}-4 x+2 y=36$$

Step-by-Step Solution

Answer

The center of the circle is (2, -1) and the radius is \( \sqrt{41} \)

Step 1: Completing the Square for x

To complete the square for the x-terms, take the coefficient of x, divide by 2, and square it. The coefficient of x is -4. Dividing by 2 gives -2 and squaring it gives 4. Add 4 to both sides of the equation to maintain equality.

Step 2: Completing the Square for y

Repeat the 'complete the square' process for the y-terms. The coefficient of y is 2. Dividing by 2 gives 1 and squaring it gives 1. Add 1 to both sides of the equation to maintain equality.

Step 3: Rewrite the Equation in Standard Form

The equation is now in a form that we can recognize the standard form of a circle. By rearranging the terms, it becomes \( (x-2)^2 + (y+1)^2 = 36 + 4 + 1 \), simplifying to \( (x-2)^2 + (y+1)^2 = 41 \).

Step 4: Identify the Center and Radius

The standard form of a circle's equation is \( (x-h)^2 + (y-k)^2 = r^2 \). By comparing, we find the center \( (h, k) \) of the circle is \( (2, -1) \) and the radius \( r \) is the square root of 41.

Step 5: Graph the Circle

Plot the center at \( (2, -1) \) on a graph. Use the radius to make a circle around the center and ensure that all points on the circle are equidistant from the center, exactly at the radius's length.

Key Concepts

Standard Form of a Circle

The standard form of a circle's equation is an essential concept in algebra and geometry, as it allows us to easily identify a circle's key attributes: its center and radius. The standard form is written as

\[ (x - h)^2 + (y - k)^2 = r^2 \]

where

• \( (h, k) \) are the coordinates of the circle's center,
• \( r \) is the radius of the circle,
• \( x \) and \( y \) are variables that represent any point \( (x, y) \) on the circle.

To transform the general form of a circle's equation into standard form, the process of 'completing the square' is used for both \( x \) and \( y \) terms. This involves adding the square of half the coefficient of \( x \) and \( y \) to both sides of the equation to create perfect square trinomials on the left side. The result is a reorganized equation reflecting the standard form, which immediately gives us access to the center and the radius upon comparison with the general standard form equation.

Circle Graphing

Graphing a circle involves translating the standard form equation into a visual representation on a coordinate plane. Here is a step-by-step process:

• Identify the center \( (h, k) \) from the standard form equation.
• Calculate the radius \( r \) by taking the square root of the value on the right side of the equation.
• Mark the point \( (h, k) \) on the graph as the center of the circle.
• Using a compass or by estimation, draw a circle around the center, with all points being \( r \) units away from \( (h, k) \).

The complete circle on a graph visually demonstrates the set of all points that are equidistant from a single point, the center. It's essential for students to practice with different equations and radii to become proficient in graphing circles accurately.

Center and Radius of a Circle

The center and the radius are the defining features of a circle. To find these, start with the circle's equation in standard form, \( (x-h)^2 + (y-k)^2 = r^2 \):

• The coordinates \( (h, k) \) of the center are found directly from the positions of \( h \) and \( k \) in this equation. If the equation is given in general form, we shift it into standard form to reveal these coordinates.
• The radius \( r \) is found by taking the square root of the constant term that resides on the right side of the standard form equation.

In the context of the provided exercise, we adapt this approach to find that the center of the circle is at \( (2, -1) \) and the radius is the square root of 41. Understanding that the center provides a fixed point from which every point on the circle is an exact radius away, and the radius offers a measurement of how large the circle is from the center is vital. These components are crucial for graphing, solving, and applying properties of circles in geometry.