Problem 15 - 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. $$ x^{2}+y^{2}-2 x+10 y+19=0 $$

Step-by-Step Solution

Answer

Circle with center (1, -5) and radius \(\sqrt{7}\).

Step 1 - Rearrange the Equation

Rewrite the equation in the standard form for a circle by completing the square. The given equation is: \[ x^{2} + y^{2} - 2x + 10y + 19 = 0 \]

Step 2 - Complete the Square for x

To complete the square for the x terms, take the coefficient of x, divide by 2, and square it: \[ (x - 1)^{2} = x^{2} - 2x + 1 \] Adjust the equation accordingly: \[ x^{2} - 2x = (x - 1)^{2} - 1 \]

Step 3 - Complete the Square for y

To complete the square for the y terms, take the coefficient of y, divide by 2, and square it: \[ (y + 5)^{2} = y^{2} + 10y + 25 \] Adjust the equation accordingly: \[ y^{2} + 10y = (y + 5)^{2} - 25 \]

Step 4 - Rewrite the Equation

Substitute the complete square forms back into the original equation: \[ (x - 1)^{2} - 1 + (y + 5)^{2} - 25 + 19 = 0 \] Simplify the resulting expression: \[ (x - 1)^{2} + (y + 5)^{2} - 7 = 0 \]

Step 5 - Identify the Circle Formula

Rewrite it in the form of a circle equation by isolating the constant term: \[ (x - 1)^{2} + (y + 5)^{2} = 7 \]

Step 6 - Interpret the Result

Recognize that this represents a circle with center (1, -5) and radius \(\sqrt{7}\).

Key Concepts

Circle Equation

A circle's equation in algebra typically appears in the form \(x^2 + y^2 + Dx + Ey + F = 0\). This general form can help us identify a circle, but it's not always easy to understand at first glance. That's why we often convert it into a more recognizable format, the standard form. In the given exercise, we start with \(x^2 + y^2 - 2x + 10y + 19 = 0\). By rearranging and simplifying, we aim to rewrite this equation in a way that makes it easier to understand as a circle equation. To do this, we use a method called 'completing the square'. This method allows us to group the x terms and y terms separately to see the circle's center and radius more clearly.

Standard Form of a Circle

The standard form of a circle equation looks like this: \((x - h)^2 + (y - k)^2 = r^2\). Here, \( (h, k)\) is the center of the circle, and \( r\) is the radius. Our goal in the exercise is to transform the given equation into this form.

Initially, we had \(x^2 + y^2 - 2x + 10y + 19 = 0\). First, we complete the square for the x terms: \(x^2 - 2x\) becomes \((x - 1)^2 - 1\). We do the same for the y terms: \(y^2 + 10y\) becomes \((y + 5)^2 - 25\).

We then substitute these forms back into the equation: \((x - 1)^2 - 1 + (y + 5)^2 - 25 + 19 = 0\). After simplifying, we get \((x - 1)^2 + (y + 5)^2 = 7\). Now, it’s in the standard form of a circle, with center \( (1, -5) \) and radius \(\root 7\).

Algebraic Manipulation

Algebraic manipulation is the process of rearranging and simplifying equations to reveal useful information. In this exercise, we performed several algebraic manipulations to complete the square and convert the given equation into the standard form of a circle.

The steps include:

• Identifying the coefficients of the x and y terms.
• Completing the square for both x and y variables by taking half of the coefficients, squaring them, and adjusting the equation accordingly.
• Substituting the squared terms back into the original equation.
• Simplifying the equation step-by-step until we get it into the standard form of a circle.

These algebraic techniques are essential in mathematics because they help transform complex-looking equations into more familiar and useful forms, aiding our understanding of the relationships they represent.