Problem 25 - Algebra A Combined Function

Question

The graph of each equation is a circle. Find the center and the radius, and then graph the circle. See Examples 5 through 7. $$ x^{2}+y^{2}+2 x-4 y=4 $$

Step-by-Step Solution

Answer

The center is (-1, 2) and the radius is 3.

Step 1: Rearrange the Equation

First, group the terms for each variable together. The original equation is \( x^2 + 2x + y^2 - 4y = 4 \).

Step 2: Complete the Square for x

To complete the square for the \(x\) terms, take the coefficient of \(x\), divide by 2, and square it. \(x^2 + 2x\) becomes \((x+1)^2 - 1\).

Step 3: Complete the Square for y

For the \(y\) terms, take the coefficient of \(y\), divide by 2, and square it. \(y^2 - 4y\) becomes \((y-2)^2 - 4\).

Step 4: Substitute and Simplify

Substitute the completed squares back into the equation: \((x+1)^2 - 1 + (y-2)^2 - 4 = 4\). Combine and simplify: \((x+1)^2 + (y-2)^2 = 9\).

Step 5: Identify the Center and Radius

The equation \((x+1)^2 + (y-2)^2 = 3^2\) is a circle with center \((-1, 2)\) and radius 3.

Key Concepts

Completing the Square

Completing the square is a fundamental technique in algebra used to transform a quadratic expression into a perfect square. This method is especially useful when dealing with conic sections, like circles, in their general form.
To complete the square, follow these simple steps:

• Identify the quadratic and linear terms in the expression for each variable.
• For a term like \(x^2 + bx\), take the coefficient of \(x\), divide it by 2, then square it. This creates the form \((x + \frac{b}{2})^2\) but remember to balance by subtracting the same squared value.
• Apply the same for terms involving \(y\), converting \(y^2 + by\) into \((y + \frac{b}{2})^2\).

In the exercise, we start with \(x^2 + 2x\). The coefficient of \(x\) is 2, divide by 2 to get 1, and square it to obtain 1. Thus, \(x^2 + 2x\) becomes \((x+1)^2 - 1\).
The process is repeated for \(y^2 - 4y\) where the coefficient is -4. Divide by 2 to get -2, square it to get 4, completing to \((y-2)^2 - 4\).
This transformation yields expressions that can be easily interpreted, enabling us to rewrite quadratic equations into recognizable forms for graphing.

Graphing Circles

Graphing a circle requires identifying key characteristics from its equation, usually in the standard form \((x-h)^2 + (y-k)^2 = r^2\).
The terms \((h, k)\) represent the center of the circle, while \(r\) is the radius. This form arises naturally once we have completed the square for both \(x\) and \(y\) terms as shown in the exercise solution.

• Consider the newly formed equation: \((x+1)^2 + (y-2)^2 = 9\).
• Here, \(x+1\) and \(y-2\) suggest adjustments to the \(x\) and \(y\) positions, indicating a center at \((-1, 2)\).
• The term \(9\) on the right is \(3^2\), pointing to a radius of 3.

To graph, plot \((-1, 2)\) as the center. From there, measure 3 units in all directions (up, down, left, right) to outline the circle. This gives a visual representation based on algebraic manipulation.

Equation of a Circle

The equation of a circle in algebra provides a neat way to visually and mathematically describe a set of points equidistant from a center point. The standard form mentioned is derived from applying algebraic techniques like completing the square.
The fundamental components of the circle’s equation \((x-h)^2 + (y-k)^2 = r^2\) are:

• \(h\) and \(k\): These represent the circle's center in the Cartesian plane. We deduced \(-1\) and \(2\) respectively from the equation \((x+1)^2 + (y-2)^2 = 9\).
• \(r\): The radius \(3\), is crucial in understanding the scale or size of the circle, calculated as the square root of the constant term on the right side, \(9\), in this case.

This formula elegantly sums up a geometric shape into an algebraic expression. When graphed, it vividly represents all points that satisfy this equation, forming a perfect circle on the plane.