Problem 16 - The Calculus with Analytic Geometry

Question

In Exercises 15 through 20 , determine whether the graph is a circle, a point- circle, or the empty set. $$ 4 x^{2}+4 y^{2}+24 x-4 y+1=0 $$

Step-by-Step Solution

Answer

The equation represents a circle with center at \((-3, \frac{1}{2})\) and radius \sqrt{10}.

Step 1: Expand and Simplify the Equation

Rewrite the given equation to a more recognizable form. Start by grouping the terms involving x and y: \[4x^2 + 24x + 4y^2 - 4y + 1 = 0\]

Step 2: Complete the Square

To complete the square, group the x and y terms separately:\[4(x^2 + 6x) + 4(y^2 - y) + 1 = 0\]Factor out the coefficient of the squared terms:\[4[(x^2 + 6x) + (y^2 - y)] + 1 = 0\]

Step 3: Further Complete the Square for x-terms

Inside the brackets for x-terms, complete the square:\[x^2 + 6x = (x + 3)^2 - 9\]Rewrite the expression:\[4[(x + 3)^2 - 9 + (y^2 - y)] + 1 = 0\]

Step 4: Complete the Square for y-terms

Inside the brackets for y-terms, complete the square:\[y^2 - y = (y - \frac{1}{2})^2 - \frac{1}{4}\]Rewrite the updated equation:\[4[(x + 3)^2 - 9 + (y - \frac{1}{2})^2 - \frac{1}{4}] + 1 = 0\]

Step 5: Simplify to get Standard Circle Form

Expand and combine constants:\[4[(x + 3)^2 + (y - \frac{1}{2})^2 - 9.25] + 1 = 0\]Simplify further:\[4(x + 3)^2 + 4(y - \frac{1}{2})^2 - 41 + 1 = 0\]\[4(x + 3)^2 + 4(y - \frac{1}{2})^2 - 40 = 0\]\[4((x + 3)^2 + (y - \frac{1}{2})^2) = 40\]Divide by 4:\( (x + 3)^2 + (y - \frac{1}{2})^2 = 10\)

Step 6: Determine the Shape

The equation \( (x + 3)^2 + (y - \frac{1}{2})^2 = 10 \) is in the standard form of a circle \((x - h)^2 + (y - k)^2 = r^2\) where \(h = -3\), \(k = \frac{1}{2}\), and \(r^2 = 10\), indicating that it represents a circle with center at \((-3, \frac{1}{2})\) and radius \( \sqrt{10} \).

Key Concepts

Completing the Square

Completing the square is a method used to transform a quadratic equation into a perfect square trinomial. This approach helps to easily convert a quadratic equation into its vertex or standard form, making it simpler to identify key features like the center and radius of a circle in analytic geometry.
To complete the square, follow these steps:

• Group the x and y terms separately.
• Factor out any coefficients of the squared terms.
• Add and subtract a suitable constant to complete the square for each group.
• Rewrite each group as a binomial squared.

In our example, we started with the equation ewline. ewline \[ 4x^2 + 24x + 4y^2 - 4y + 1 = 0 \] ewline We grouped the x and y terms: ewline \[ 4(x^2 + 6x) + 4(y^2 - y) + 1 = 0 \] ewline By factoring out the 4 from each group and completing the square: ewline \[ 4((x + 3)^2 - 9 + (y - \frac{1}{2})^2 - \frac{1}{4}) + 1 = 0 \] ewline This transformation helped us transform the quadratic equation into a more manageable form.

Standard Form of a Circle

The standard form of a circle's equation is ewline \[ (x - h)^2 + (y - k)^2 = r^2 \] ewline Here, (h, k) represents the center of the circle, and r is the radius. ewlineTo get the standard form from a general quadratic equation, you perform steps like completing the square. Once complete, you can derive the circle's center and radius quickly. ewlineIn the provided example, by performing and simplifying the squares, we arrived at ewline \[ (x + 3)^2 + (y - \frac{1}{2})^2 = 10 \] ewline Here, it is clear that: ewline

• The center (h, k) is (-3, \frac{1}{2})
• The radius r is ewline\[ \frac{1}{2} \]

This easy-to-read form enables immediate understanding of the circle's geometry.

Analytic Geometry

Analytic geometry uses algebraic equations to represent geometric shapes, allowing for the solving of geometry problems using algebraic techniques.
In the context of circles, this discipline helps in transforming equations and extracting geometric properties such as the center and radius of a circle from algebraic expressions.
Key concepts include:

• Using the distance formula to find the radius.
• Applying coordinate transformations to standardize equations.
• Understanding the intersection of geometric shapes through equations.

For our example, stepping through the equation ewline \[4x^2 + 24x + 4y^2 - 4y + 1 = 0 \], ewline we used completing-the-square and reduced it to the standard form ewline \[(x + 3)^2 + (y - \frac{1}{2})^2 = 10 \] ewline In analytic geometry, this transformation allows for a clear identification of the circle's center and radius. The combination of algebra and geometry makes solving complex problems more approachable and understandable.