Problem 22 - Algebra for College Students

Question

For Problems \(15-32\), find the center and the length of a radius of each of the circles. $$ x^{2}+y^{2}-6 x-10 y+30=0 \quad(3,5), r=2 $$

Step-by-Step Solution

Answer

The center is (3, 5) and the radius is 2.

Step 1: Identify the Circle Equation

Given the equation of the circle: \(x^2 + y^2 - 6x - 10y + 30 = 0\). This equation is in the general form \(x^2 + y^2 + Dx + Ey + F = 0\).

Step 2: Move Constant to Other Side

Rearrange the equation to isolate the constant term on the right-hand side: \(x^2 + y^2 - 6x - 10y = -30\).

Step 3: Complete the Square for x-terms

For the x-terms: \(x^2 - 6x\), complete the square by taking half of the coefficient of \(x\), which is \(-6\), dividing it by 2 to get \(-3\), and then squaring it to get \(9\). Add and subtract \(9\) inside the equation. This gives \((x-3)^2 - 9\).

Step 4: Complete the Square for y-terms

For the y-terms: \(y^2 - 10y\), complete the square by taking half of \(-10\), dividing by 2 to get \(-5\), and squaring it to get \(25\). Add and subtract \(25\) inside the equation. This gives \((y-5)^2 - 25\).

Step 5: Simplify the Equation

Substitute the completed squares into the equation \((x-3)^2 - 9 + (y-5)^2 - 25 = -30\), then simplify it to \((x-3)^2 + (y-5)^2 = 4\).

Step 6: Identify Center and Radius

Recognize the standard form of a circle \((x-h)^2 + (y-k)^2 = r^2\), which gives the center \((h, k)\) and the radius \(r\). From the equation \((x-3)^2 + (y-5)^2 = 4\), the center is \((3, 5)\) and \(r^2 = 4\) so \(r = 2\).

Key Concepts

Completing the Square

Completing the square is a mathematical technique that makes it easier to rewrite quadratic equations, which often helps in solving them or transforming them into a more useful form. In terms of circles, completing the square aids in converting the equation of a circle from its general form to its standard form.

Imagine you have the equation of a circle like this: \[x^2 + y^2 - 6x - 10y + 30 = 0.\] In this context, completing the square is your tool to rearrange and regroup the terms involving variables like "x" and "y" separately.

Here’s how you do it step-by-step:

• Look at the "x" terms, \(x^2 - 6x\). Take half of the coefficient of x, which is \(-6\), resulting in \(-3\), and square it, resulting in 9. Add and subtract 9 to partially complete the square.
• Do a similar process for the "y" terms, \(y^2 - 10y\). Half of \(-10\) is \(-5\), squared is 25. Again, add and subtract 25.
• The equation is now easier to handle and looks like: \((x-3)^2 - 9 + (y-5)^2 - 25 = 0.\)

Now, bring the numbers \(-9\) and \(-25\) to the other side to simplify further, which leads to the standard form of the circle equation.

Geometry

Geometry, the branch of mathematics that deals with the shapes, sizes, and properties of figures, plays a critical role in understanding, any geometric forms, such as circles.

The equation of a circle, in particular, represents a set of points in a plane that are all equidistant from a center point. Knowing how to interpret and manipulate this equation is key to understanding the geometric properties of the circle. For example, this relates to the real world when creating designs or solving practical problems where circular patterns are involved.

In this exercise, when transforming the general equation using completing the square, you're implementing geometric concepts to identify key circle characteristics: namely, its center and radius. This is paramount for visualizing and planning layouts in both simple and complex designs.

Radius Calculation

The radius of a circle is the distance from the center to any point on the circle. Knowing the radius is crucial, especially in practical applications such as engineering and construction, where precision is paramount.

In the standard equation of a circle: \[(x-h)^2 + (y-k)^2 = r^2,\]"\((h, k)\)" represents the center of the circle, and \(r^2\) is the square of the radius. After completing the square, we derived the equation: \((x-3)^2 + (y-5)^2 = 4.\)By recognizing it in this standard form, it becomes clear that \(r^2 = 4\). Therefore, the actual radius \(r\) is the square root of 4, which is 2.
This essential understanding of how to extract "\(r\)" from this equation is foundational to working with circles and is a simple yet powerful tool.

Standard Form of a Circle

The standard form of a circle's equation is a transformational toolkit for clarity and insight when dealing with circular equations. It provides a streamlined way to identify a circle’s geometric features.

Written mathematically as: \[(x-h)^2 + (y-k)^2 = r^2,\]this form reveals the center \((h, k)\) and the radius \(r\) directly from the equation.

To illustrate, when we transform the original equation to: \((x-3)^2 + (y-5)^2 = 4\), we can immediately determine the circle's center is at \((3,5)\), and the radius squared is 4. Consequently, the radius is 2.

Understanding how and why to rearrange a circle equation into this neat form unlocks an easier path to solving geometric problems and appreciating the beauty inherent in these symmetrical forms. This form is not just a mathematical convenience but a visual aid, perfect for applications like plotting points on a graph or designing components in physical projects.