Problem 23 - Algebra and Trigonometry Real Mathematics, Real People

Question

Write the equation of the circle in standard form. Then identify its center and radius. $$x^{2}+y^{2}-2 x+6 y+9=0$$

Step-by-Step Solution

Answer

The equation of the circle in standard form is \((x-1)^2+(y+3)^2 = 10\), the center of the circle is (1, -3) and its radius is approximately 3.162.

Step 1: Rewrite in completed square form

Firstly, group the x's and y's together. This results in the equation \((x^{2}-2x)+(y^{2}+6y) = -9\). Then complete the square for each grouping by adding the square of half the coefficient of x in the x's grouping and the square of half the coefficient of y in the y's grouping to both sides of the equation. This results in \((x-1)^2+(y+3)^2 = 1^2+3^2=10\).

Step 2: Identify the center and radius of the circle

Now in \((x-1)^2+(y+3)^2 = 10\), one can see that the circle's center (h, k) is at (1, -3) and the radius r is the square root of 10, or approximately 3.162.

Key Concepts

Completing the Square

Completing the square is a mathematical technique used to transform a quadratic expression into a perfect square trinomial. This technique is particularly useful in converting the general form of a quadratic equation into the standard form, making it easier to graph or solve. For our circle equation,

• Start by grouping terms: for example, \( (x^2 - 2x)\) and \((y^2 + 6y) \) in the exercise.
• To complete the square for \(x^2 - 2x\), take half of the linear coefficient \(-2\), square it to get \(1\), and add it inside the parentheses to form \((x-1)^2\).
• Similarly, for \( y^2 + 6y \), half is \(3\), squared is \(9\), leading to \((y+3)^2\).
• Add the square values to both sides of the equation to maintain balance.

This method reorganizes the original quadratic into a form \((x-a)^2 + (y-b)^2\), perfectly setting the stage for identifying the circle's characteristics.

Standard Form of a Circle Equation

The standard form of a circle's equation is expressed as:\[(x-h)^2 + (y-k)^2 = r^2\]where

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

This form provides a direct way to visualize and understand the circle's geometry. By converting an equation to this standard form, like in our exercise result \((x-1)^2 + (y+3)^2 = 10\), we can instantly see that:

• The center is at \((1, -3)\)
• The radius, calculated from the square root of 10, is approximately \(3.162\).

The equation clearly communicates all essential aspects of the circle, simplifying graphing and analysis.

Radius and Center of a Circle

The radius and center are fundamental properties defining a circle. In any circle equation in standard form, these properties are straightforward to identify:

• The center occurs at point \((h, k)\) in the equation \((x-h)^2 + (y-k)^2 = r^2\).
• The radius \(r\) is the distance from the center to any point on the circle, found as the square root of \(r^2\).

From our example, \((x-1)^2 + (y+3)^2 = 10\),:

• The center is at \((1, -3)\)
• The radius, \(\sqrt{10}\), approximately \(3.162\).

Understanding these attributes facilitates the graphing process and deepens comprehension of the circle's layout within a plane.

Conic Sections

A circle is one of the conic sections, which also include ellipses, parabolas, and hyperbolas. These curves are derived by intersecting a plane with a double-napped cone in various angles and positions.

• A circle results when the intersecting plane is perpendicular to the cone's axis, creating a perfectly round boundary.
• The simplicity of a circle makes it a fundamental geometric shape present in many mathematical applications.

Understanding circles within the context of conic sections helps place them within the broader study of geometry and emphasizes their unique properties compared to other conics like ellipses or parabolas.