Problem 24 - Algebra for College Students

Question

For Problems \(15-32\), find the center and the length of a radius of each of the circles. $$ x^{2}+y^{2}+10 x+14 y+73=0 \quad(-5,-7), r=1 $$

Step-by-Step Solution

Answer

Center: (-5, -7), Radius: 1

Step 1: Recognize the Equation Form

The given circle equation is \(x^{2} + y^{2} + 10x + 14y + 73 = 0\). This is a non-standard form of a circle equation, and we need to rewrite it to identify the center and radius.

Step 2: Convert to Standard Form

Rewrite the equation by completing the square for both \(x\) and \(y\):1. Group the \(x\) terms: \(x^{2} + 10x\)2. Group the \(y\) terms: \(y^{2} + 14y\)3. Move the constant to the other side of the equation: \(x^{2} + 10x + y^{2} + 14y = -73\)

Step 3: Complete the Square for \(x\)

Take the coefficient of \(x\) (which is 10), halve it to get 5, then square it (5² = 25):Add and subtract 25 from the equation: \((x^{2} + 10x + 25) - 25\)

Step 4: Complete the Square for \(y\)

Take the coefficient of \(y\) (which is 14), halve it to get 7, then square it (7² = 49):Add and subtract 49 from the equation: \((y^{2} + 14y + 49) - 49\)

Step 5: Rearrange the Equation

Combine the completed squares and the constant terms:\((x + 5)^2 + (y + 7)^2 = -73 + 25 + 49 = 1\)

Step 6: Identify the Center and Radius

The equation is now in the standard circle form: \((x - h)^2 + (y - k)^2 = r^2\), where the center is \((h, k)\) and the radius is \(r\).From the equation \((x + 5)^2 + (y + 7)^2 = 1\):- Center \((h, k)\) is \((-5, -7)\)- Radius \(r\) is \(\sqrt{1} = 1\)

Key Concepts

Standard Form of Circle

The standard form of a circle equation is a specific way of writing the equation that clearly reveals the circle's center and radius. The form looks like this: \[(x - h)^2 + (y - k)^2 = r^2\]Here:

• \(h\) and \(k\) represent the coordinates of the circle's center, \((h, k)\).
• \(r\) is the radius of the circle.

To transform an equation into this form, we often have to rearrange it and perform operations such as completing the square, to isolate these variables. Being comfortable with this form allows you to immediately identify the circle's center and size just by looking at the equation.

Completing the Square

Completing the square is a technique often used to rewrite a quadratic equation in a simple form that makes it easy to read off the properties of a graph, such as the center and radius of a circle. For a term like \(x^2 + bx\), we complete the square by:

• Halving the \(b\), value giving us \(\frac{b}{2}\).
• Squaring this result for \((\frac{b}{2})^2\).
• Adding and subtracting \((\frac{b}{2})^2\) to the equation.

This process forms a perfect square trinomial, which can be expressed as \((x + \frac{b}{2})^2\). By using this skill, you can convert messy circle equations into the neat standard form. In our exercise, it was used to transform the \(x^2 + 10x\) and \(y^2 + 14y\) terms.

Circle Center

The center of a circle, denoted as \((h, k)\), is the point that is equidistant from all points on the circumference. When the circle's equation is in its standard form as \((x - h)^2 + (y - k)^2 = r^2\), the values of \(h\) and \(k\) directly tell us the circle’s center.Finding the center is pivotal when sketching or understanding the circle’s position within its plane. In the given exercise, after completing the square, we derived the equation \((x + 5)^2 + (y + 7)^2 = 1\). From this, it's clear that the center is at \((-5, -7)\). This central point forms the baseline for understanding how the circle maps onto a graph.

Circle Radius

The radius of a circle is the measure of distance from its center to any point on its edge. It is a crucial aspect of the circle, reflecting its size. In the standard circle equation, \((x - h)^2 + (y - k)^2 = r^2\), the radius \(r\) is derived from \(r^2\).To extract the radius:

• Observe the number on the right side of the equation, which is \(r^2\).
• Take the square root of this number to find \(r\).

In our problem, we ended up with the equation \((x + 5)^2 + (y + 7)^2 = 1\), indicating that \(r^2 = 1\). Thus, the radius is \(r = \sqrt{1} = 1\). This small radius shows that the circle is quite compact, centered around the designated point.