Problem 58 - College Algebra

Question

Complete the square and write the equation in standard form. Then give the center and radius of each circle and graph the equation. $$ x^{2}+y^{2}+12 x-6 y-4=0 $$

Step-by-Step Solution

Answer

The given equation in standard form is \((x + 6)^{2} + (y - 3)^{2} = 49\). The center of the circle is \((-6, 3)\) and the radius of the circle is \(7\).

Step 1: Rearranging the terms

First rearrange the equation by grouping x terms and y terms together: \( (x^{2} + 12x) + (y^{2} - 6y) = 4 \)

Step 2: Complete the square for x and y

Completing the square for \(x^{2} + 12x\), a term \((b/2)^2\) is added and subtracted within the bracket where \(b\) is the coefficient of \(x\). Here, \(b\) is \(12\), so the term added and subtracted will be \((12/2)^2\) which is \(36\). Similarly for \(y^{2} - 6y\), \(((-6)/2)^2\) i.e. \(9\) will be added and subtracted in the bracket. Putting these calculations in the original equation gives: \( (x^{2} + 12x + 36) - 36 + (y^{2} - 6y + 9) - 9 = 4 \)

Step 3: Simplify the equation

Now simplify above expression you get, \((x+6)^{2} - 36 + (y-3)^{2} - 9 = 4\). Further simplifying and arranging to form of a circle equation, you get \((x + 6)^{2} + (y - 3)^{2} = 49 \)

Step 4: Identify the centre and the radius of the circle

On comparing the resulting equation with equation of circle which is \((x - h)^{2} + (y - k)^{2} = r^{2}\), we can identify \(h = -6, k = 3\) and \(r = 7\). So, the circle is centered at \((-6,3)\) and has a radius \(7\)

Key Concepts

Standard Form of a Circle

Understanding the standard form of a circle's equation is essential for solving a variety of problems in geometry and algebra. It presents a way to easily identify a circle's key properties: its center and radius. The standard form of a circle is given by the equation \( (x - h)^{2} + (y - k)^{2} = r^{2} \), where \( (h, k) \) are the coordinates of the circle's center and \( r \) is its radius.

In this exercise, completing the square transforms the given general equation of a circle into its standard form. This technique organizes the equation so that x and y components are squared and isolated, allowing the identification of \( h \) and \( k \) along with determining the value for \( r^{2} \). By reformatting the equation, we can effortlessly extract the circle's defining features.

Circle Graphing

Graphing a circle involves plotting it on a coordinate plane with the correct center and radius as established from its standard form equation. Equipped with the center point \( (h, k) \) and radius \( r \) of the circle, you can sketch the circle by marking the center and moving \( r \) units away from it in all directions to define the circumference.

For the equation in the given exercise, after completing the square and rearranging into standard form, we determine the center at \( (-6,3) \) and the radius as \( 7 \) units. On a graph, we start at point \( (-6,3) \) and measure 7 units up, down, left, and right to establish points on the circle, then draw a smooth curve connecting those points to represent the circle's edge. Graphing circles like this enhances understanding of their spatial properties and visualizes their geometric relations on the plane.

Center and Radius of a Circle

The center and radius of a circle are foundational concepts in its geometry. The center is the equidistant point from all points on the circle's edge, and the radius is that constant distance. In the equation's standard form, these values can be directly read off as the variables \( h \) and \( k \) for the center \( (h, k) \) and \( r \) for the radius.

For our exercise, after rewriting the given equation as \( (x + 6)^{2} + (y - 3)^{2} = 49 \) through completing the square, it becomes clear that the centroidal coordinates are \( h = -6 \) and \( k = 3 \) by comparing the equation to the standard form \( (x - h)^{2} + (y - k)^{2} = r^{2} \). Additionally, we identify the radius as \( r = 7 \) because \( r^{2} = 49 \). Understanding the center and radius allows us to accurately graph the circle and apply various analytical methods when solving related geometric problems.