Problem 51 - Calculus Single Variable

Question

Which of the equations are circles? Which are not? Give precise reasons for your answers. \(5 x^{2}+5 y^{2}=2 x-3 y+6\)

Step-by-Step Solution

Answer

The equation represents a circle because it can be rearranged to match a circle's standard form.

Step 1: Rearrange the Equation

Start by rearranging the given equation to identify its standard form. Given equation is: \[5x^2 + 5y^2 = 2x - 3y + 6\] Move all terms to one side to form:\[5x^2 + 5y^2 - 2x + 3y - 6 = 0\]

Step 2: Group and Simplify Terms

Next, group the terms with similar variables. Divide through by 5 to simplify the equation:\[x^2 + y^2 - \frac{2x}{5} + \frac{3y}{5} = \frac{6}{5}\]This makes it easier to identify any resemblance to the standard form of a circle's equation.

Step 3: Identify Circle Equation Format

Identify whether the simplified equation maintains the format of a circle equation. A circle's equation in standard form is \[(x - h)^2 + (y - k)^2 = r^2\]To check this, observe if the coefficients of \(x^2\) and \(y^2\) are equal and the linear terms can form perfect squares.

Step 4: Assess Equation for Circle Characteristics

The simplified equation now shows equal coefficients for \(x^2\) and \(y^2\), both equal to 1, which aligns with a circle's form before completing the square. Add that the linear terms \(-\frac{2x}{5}\) and \(+\frac{3y}{5}\) could be integrated into perfect squares, confirming circle characteristics.

Key Concepts

Standard Form of a Circle

The standard form of a circle's equation is a special way to write the circle's equation so that its properties are easily recognizable. It looks like this:\[ (x - h)^2 + (y - k)^2 = r^2 \] In this equation,

• \( (h, k) \) is the center of the circle
• \( r \) is the radius of the circle

To transform any circle equation into its standard form, it's crucial to make sure that:- The coefficients of \( x^2 \) and \( y^2 \) are the same.- The expression is structured such that it matches the perfect square format for \( (x - h)^2 \) and \( (y - k)^2 \). Breaks in this order suggest the equation doesn't precisely represent a standard circle.For example, in the given equation \( x^2 + y^2 - \frac{2x}{5} + \frac{3y}{5} = \frac{6}{5} \), we notice the equal coefficients for \( x^2 \) and \( y^2 \). Each coefficient (here, both are 1) hints at the symmetry necessary for forming a circle.

Completing the Square

Completing the square is the process that helps us convert a quadratic equation into a format that highlights its geometric shape—like turning a general quadratic equation into the standard circle form. This method is helpful in identifying the radius and the center of the circle from the linear terms.Here's how you complete the square:

• For a term like \( x^2 - bx \), add and subtract \( \left(\frac{b}{2}\right)^2 \). This changes the equation:

\[ x^2 - bx + \left(\frac{b}{2}\right)^2 - \left(\frac{b}{2}\right)^2 \] To a perfect square: \[ \left(x - \frac{b}{2}\right)^2 \] With our equation, to complete the square for \(-\frac{2x}{5}\) and \(+\frac{3y}{5}\), first divide and adjust to create perfect squares. This structural shift reveals valuable circle properties—like its center and radius—concealed until rearranged.

Circle Characteristics

The characteristics of a circle start with understanding its basic components: the center and the radius. Each component is anchored in the equation's layout.

• The **center** is defined by \( (h, k) \) in the equation \( (x - h)^2 + (y - k)^2 = r^2 \). Thus, understanding and forming the perfect square in the equation guides you directly to this central point.
• The **radius** is straightforward—it is \( r \) and square rooted from the right side of the equation. It scales the circle, telling us how wide it spans from its center.

Interpreting such characteristics ensures clarity. For instance, a differentiator for our original equation modelled itself around identifying equality in \( x^2 \) and \( y^2 \) coefficients, setting the groundwork for spherical symmetry and validating its classification as a circle. Keep these traits in mind when analyzing any circle equation—it'll be easier to spot and understand its nuances.
Remember: Observing the balance in an equation and formulating it into squares enriches your geometry toolkit.