Problem 69 - Intermediate Algebra

Question

Sketch the graph of each equation. If the graph is a parabola, find its vertex. If the graph is a circle, find its center and radius. $$x^{2}+y^{2}+2 x+12 y-12=0$$

Step-by-Step Solution

Answer

Center: (-1, -6); Radius: 7

Step 1: Rewrite the Equation by Rearranging Terms

The given equation is \(x^2 + y^2 + 2x + 12y - 12 = 0\). The terms are already arranged with all the quadratic terms grouped together. However, to complete the squares, we will first group the \(x\) and \(y\) terms separately: \[(x^2 + 2x) + (y^2 + 12y) = 12\]

Step 2: Complete the Square for x-Terms

Focus on the \(x\)-terms \(x^2 + 2x\). Add and subtract \(1\) (since \((\frac{2}{2})^2 = 1\)) to complete the square: \[(x^2 + 2x + 1 - 1) = (x+1)^2 - 1\] Replace \(x^2 + 2x\) in the equation with \((x+1)^2 - 1\).

Step 3: Complete the Square for y-Terms

Now focus on the \(y\)-terms \(y^2 + 12y\). Add and subtract \(36\) (since \((\frac{12}{2})^2 = 36\)) to complete the square:\[(y^2 + 12y + 36 - 36) = (y+6)^2 - 36\]Replace \(y^2 + 12y\) in the equation with \((y+6)^2 - 36\).

Step 4: Form the Completed Square Equation

Substituting the completed squares into the main equation, we have: \[(x+1)^2 - 1 + (y+6)^2 - 36 = 12\]Simplify it: \[(x+1)^2 + (y+6)^2 = 49\]

Step 5: Identify the Center and Radius of the Circle

The equation \((x+1)^2 + (y+6)^2 = 49\) is in the standard form of a circle \((x-h)^2 + (y-k)^2 = r^2\), where \((h,k)\) is the center and \(r\) is the radius. Comparing, we find the center to be \((-1, -6)\) and the radius \(r\) is \(7\), since \(\sqrt{49} = 7\).

Key Concepts

Completing the Square

Completing the square is an essential algebraic technique used to transform quadratic expressions into a form that reveals vital information, such as a vertex or a circle's center and radius. It involves making a quadratic expression a perfect square trinomial, which is easier to work with. Here's a breakdown of how this process works:

• For expressions like \(x^2 + bx\), we can complete the square by adding and subtracting \((\frac{b}{2})^2\). This transforms the expression into \((x+\frac{b}{2})^2-(\frac{b}{2})^2\).
• This technique also applies to polynomials with terms in \(y\), like \(y^2 + cy\). Here, add and subtract \((\frac{c}{2})^2\), resulting in \((y+\frac{c}{2})^2-(\frac{c}{2})^2\).

In our exercise, completing the square was applied separately to both the \(x\) and \(y\) terms to transform the given quadratic equation into a recognizable form of a circle equation. This makes it straightforward to spot the circle's center and radius, crucial for graphing.

Standard Form of a Circle

The standard form of a circle's equation helps us quickly determine the circle's specific properties. It's written as \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) is the circle's center and \(r\) is the radius.

There are several advantages to using this form:

• It clearly showcases the circle's center, \((h, k)\), simply from the equation's constants.
• The radius \(r\) is found by taking the square root of the constant on the right-hand side.

For instance, from the exercise's finalized equation, \((x+1)^2 + (y+6)^2 = 49\), we immediately identify the center as \((-1, -6)\). The constant \(49\) on the right side tells us the square of the radius, giving us \(r = 7\). Thus, we have all the information needed to sketch the circle accurately on a graph.

Center and Radius of a Circle

Finding the center and radius of a circle from its equation is crucial to understanding its geometric properties. These elements are vital for graphing and understanding the circle's position and size.

Here's how you can find these from the standard form equation:\

• The center, \((h,k)\), is directly taken from the equation as \((x-h)^2 + (y-k)^2 = r^2\). You adjust the signs from the expression: if you see \((x+1)\), the \(h\) coordinate is \(-1\), as deduced from \((x-h)\).
• For the radius, you look at \(r^2\) on the right side of the equation. Taking the square root gives the actual radius.

In our worked through example, converting \(x^2+y^2+2x+12y-12=0\) into \((x+1)^2 + (y+6)^2 = 49\) using completing the square shows the center is at \((-1, -6)\) with a radius of \(7\). Knowing these helps in constructing an accurate circle on a coordinate plane.