Problem 33 - Calculus with Concepts in Calculus

Question

Sketch the graph of the given equation with the help of a suitable translation. Show both the \(x\) and \(y\) axes and the \(X\) and \(Y\) axes. $$ x^{2}+y^{2}+4 x-6 y+13=0 $$

Step-by-Step Solution

Answer

The graph is a point at (-2, 3).

Step 1: Identify the Original Equation

The given equation is \(x^2 + y^2 + 4x - 6y + 13 = 0\). This is the equation of a circle that needs to be rewritten to understand its components and transformations.

Step 2: Complete the Square for x

To complete the square for \(x\), take the coefficient of \(x\), which is 4, halve it to get 2, and square it to get 4. Add and subtract 4 inside the equation for the \(x\) terms: \[ (x^2 + 4x) \rightarrow (x + 2)^2 - 4 \]

Step 3: Complete the Square for y

For \(y\), take the coefficient of \(y\), which is -6, halve it to get -3, and square it to get 9. Add and subtract 9 inside the equation for the \(y\) terms: \[ (y^2 - 6y) \rightarrow (y - 3)^2 - 9 \]

Step 4: Rewrite the Equation

Substitute the completed squares back into the equation:\[ (x + 2)^2 - 4 + (y - 3)^2 - 9 + 13 = 0 \] Simplify the equation:\[ (x + 2)^2 + (y - 3)^2 = 0 \]

Step 5: Identify the Translations

The equation \((x + 2)^2 + (y - 3)^2 = 0\) indicates a circle with center \((-2, 3)\) and radius \(0\) (a point circle). This means the graph translates the original circle's center from the origin to \((-2, 3)\).

Step 6: Sketch the Graph

Since the radius is 0, the graph consists of the single point \((-2, 3)\). On a coordinate plane, plot both the original axes (\(x\) and \(y\)) and translated axes (\(X = x + 2\) and \(Y = y - 3\)), marking the point \((-2, 3)\) as the center of the circle which is a single point.

Key Concepts

Equation of a Circle

When we talk about the equation of a circle, we're diving into one of the fundamental aspects of geometry. At its core, the standard form of a circle's equation is \( (x - h)^2 + (y - k)^2 = r^2 \), where

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

For an equation to describe a circle, the expression should fit this format. In this case, our original equation was \( x^2 + y^2 + 4x - 6y + 13 = 0 \). Notice, initially, it doesn't appear in the standard form.Ultimately, our goal is to convert this general form into the standard form. Taking into account both the \(x\) and \(y\) components, this transformation helps in identifying the circle's key characteristics: its center and radius. Once transformed, it becomes easier to interpret and plot the circle on a graph.

Completing the Square

Completing the square is a crucial algebraic technique often used to transform quadratic expressions into a form that's easier to interpret. In the context of circles, it helps us reframe the equation into a standard form. Let's break down how it's done:For the \(x\)-terms in our equation, we see \(x^2 + 4x\). Here's how you complete the square:

• Take the coefficient of \(x\) (which is 4),
• Halve it (resulting in 2),
• Then square it (getting 4).

Add and subtract this square inside the expression to balance it. This process is mirrored for the \(y\)-terms: For \(y^2 - 6y\), take \(-6\), halve to get \(-3\), and square it for 9.These steps translate the terms into perfect square trinomials. They look like this: \((x + 2)^2 - 4\) and \((y - 3)^2 - 9\). Combining them into the circle equation adds clarity and simplifies the task of identifying the circle's characteristics.

Graph Translation

Graph translation involves shifting the entire graph of an equation to a new position without altering its shape. Once we have the circle's equation in a perfect square form, identifying translations becomes a breeze.In the equation \((x + 2)^2 + (y - 3)^2 = r^2\), notice that the transformations \(x + 2\) and \(y - 3\) suggest shifts.

• The \(+2\) with \(x\) means the graph shifts left by 2 units.
• The \(-3\) with \(y\) indicates a shift upwards by 3 units.

Such transformations adjust the circle’s center from the origin to \((-2, 3)\). When plotted, the original axes (\(x\) and \(y\)) can be imagined as new axes (\(X = x + 2\) and \(Y = y - 3\)). For our specific example, since the radius is \(0\), this translates into a single point; namely the center of what would have been a circle, marked at \((-2, 3)\) on the graph.