Problem 58 - Calculus Single Variable

Question

For what values of \(k\) is the graph of \(x^{2}-8 x+y^{2}+2 y=k\) empty?

Step-by-Step Solution

Answer

The graph is empty for \(k < -17\).

Step 1: Equation of a Circle

The given equation is \(x^2 - 8x + y^2 + 2y = k\). This is the equation of a circle, although not in the standard form. Let's rewrite it to identify the center and radius of the circle.

Step 2: Complete the Square for x

Rewrite \(x^2 - 8x\) by completing the square: \(x^2 - 8x = (x-4)^2 - 16\). This transforms the x-part to \((x-4)^2 - 16\).

Step 3: Complete the Square for y

Rewrite \(y^2 + 2y\) by completing the square: \(y^2 + 2y = (y+1)^2 - 1\). This transforms the y-part to \((y+1)^2 - 1\).

Step 4: Substitute Completed Squares into Equation

Substitute the completed squares back into the original equation: \((x-4)^2 - 16 + (y+1)^2 - 1 = k\).

Step 5: Simplify the Equation

Simplify the modified equation: \((x-4)^2 + (y+1)^2 = k + 17\). This shows it in the standard circle form \((x-h)^2 + (y-k)^2 = r^2\) with center (4, -1) and radius \(\sqrt{k+17}\).

Step 6: Determine Conditions for an Empty Graph

For a circle to be empty (no real points), the radius must be negative, which occurs when \(k + 17 < 0\). Solve for \(k\): \(k < -17\). Thus, the graph is empty when \(k < -17\).

Key Concepts

Completing the Square

Completing the square is a technique used to simplify quadratic expressions. It transforms a quadratic expression into a perfect square trinomial, which makes it easier to analyze or solve.
The technique is useful in many areas, including solving quadratic equations, and, as in this context, converting equations into easily recognizable forms.
To complete the square, follow these steps:

• Identify the quadratic term and the coefficient of the linear term. For example, in the expression \(x^2 - 8x\), the quadratic term is \(x^2\) and the linear term coefficient is \(-8\).
• Take half of the coefficient of the linear term, square it, and add and subtract this value from the expression. For example, half of \(-8\) is \(-4\), and squared is \(16\). So, \(x^2 - 8x\) becomes \((x-4)^2 - 16\).
• The same process can be applied to terms in \(y\)- variable, such as \(y^2 + 2y\), where half of \(2\) is \(1\), squared is \(1\), transforming it into \((y+1)^2 - 1\).

This method transforms quadratic terms into a form that can easily be manipulated into a standard form of a circle equation.

Standard Form of a Circle

The standard form of a circle's equation is a neat way to encapsulate all the essential information about a circle at a glance. It is written as:\[(x-h)^2 + (y-k)^2 = r^2\]Here:

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

This compact form allows us to quickly determine these critical circle properties.To convert an equation into this standard form, you may need to complete the square for both \(x\) and \(y\) terms, as demonstrated in the given example. Once you complete the square, you rearrange the equation to reflect one squared term for \(x\) and one for \(y\), set equal to \(r^2\). The values obtained through completing the square provide the circle's center and equation.

Center and Radius of a Circle

The center and radius are fundamental aspects of a circle's geometry. They are straightforward to determine once a circle's equation is in the standard form.After transforming an equation like \(x^2 - 8x + y^2 + 2y = k\) into its standard form, \((x-h)^2 + (y-k)^2 = r^2\), the values \(h\) and \(k\) in the equation give you the center

• In this context, the center can be easily read off as the coordinates \((4, -1)\).

The radius is determined by taking the square root of the constant on the right-hand side of the equation:

• In the step-by-step solution, this value transforms into \(\sqrt{k+17}\), indicating that the radius depends directly on \(k\).

Understanding the relationship between the center, radius, and the standard circle form simplifies analyzing and solving circle-related problems, including determining conditions when a circle does not exist, like when the radius is negative, indicating no real solutions, giving the condition \(k < -17\) in the provided example.