Problem 58 - College Algebra Essentials

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

Step 1: Rearrange the equation

Rearrange the given quadratic equation so that all terms containing \(x\) and \(y\) are grouped together, which will allow for easier factoring. The equation becomes \((x^2 + 12x) + (y^2 - 6y) - 4 = 0\).

Step 2: Complete the square

Complete the square for each of the quadratic expressions by adding (\(b/2)^2\) to both sides of the equation, where \(b\) is the coefficient of the linear term (i.e., the term with \(x\) or \(y\)). This gives \((x^2 + 12x + 36) + (y^2 - 6y + 9) = 4 + 36 + 9 \rightarrow (x+6)^2 + (y - 3)^2 = 49\).

Step 3: Identify the center and radius

The center of the circle is given by the equations \(h = -b/2a\) and \(k = -d/2c\) where \(a, b\) are the coefficients of the \(x\) terms and \(c, d\) are the coefficients of the \(y\) terms in the standard form of the circle equation. So, \(h = -6\), \(k = 3\). The radius of the circle is the square root of the constant on the right side of the equation, so the radius is \(r = \sqrt{49} = 7\). Therefore, the center is \(-6, 3\) and the radius is 7.

Step 4: Write the standard form of the equation

The standard form of the circle equation is \((x - h)^2 + (y - k)^2 = r^2\), which in our case becomes \((x + 6)^2 + (y - 3)^2 = 49\) since \(h = -6, k = 3, \text{and} r = 7\).

Key Concepts

Completing the Square

Completing the square is a method used to convert a quadratic equation into a perfect square trinomial, which makes solving and analyzing the equation easier. In the context of circle equations, completing the square helps to identify the circle's center and radius.

To complete the square for an equation like \(x^2 + 12x\), you follow these steps:

• Take the coefficient of the linear term, which is 12 in this case, and divide it by 2: \(12/2 = 6\).
• Square the result: \(6^2 = 36\).
• Add and subtract this square inside the quadratic expression: \(x^2 + 12x + 36 - 36\).

By following this process for both the \(x\) and \(y\) terms separately, the equation becomes much easier to handle, leading to a clean standard form.

Standard Form

The standard form for the equation of a circle is given by: \[(x - h)^2 + (y - k)^2 = r^2\]where \((h, k)\) is the center of the circle and \(r\) is the radius.

This form is incredibly useful because it directly shows the circle's geometric features. After completing the square on the original equation \[x^2 + y^2 + 12x - 6y - 4 = 0\], we transformed it into the standard form \[(x + 6)^2 + (y - 3)^2 = 49\].

This makes it clear that the circle has its center at \((-6, 3)\) and a radius of \(7\). Writing equations in this form simplifies the interpretation of the circle's properties and makes graphing straightforward.

Center and Radius

Identifying the center and radius of a circle from its equation is straightforward once the equation is in standard form \[(x - h)^2 + (y - k)^2 = r^2\].

From the complete square form, \[(x + 6)^2 + (y - 3)^2 = 49\], we can deduce:

• Center \((h, k)\) is \((-6, 3)\).
• Radius \(r\) is the square root of the right side, here \(\sqrt{49} = 7\).

Knowing these values helps you visualize and graph a circle quickly. Knowing the center tells you where the circle is located on a coordinate plane, and the radius describes how large the circle is.

Graphing Circles

Graphing circles involves plotting the center on the coordinate plane and using the radius to mark the distance from the center to any point on the circle's circumference.

To graph the circle of the equation \[(x + 6)^2 + (y - 3)^2 = 49\]:

• First, plot the center at \((-6, 3)\).
• Then, from this center point, measure 7 units outward in all directions to create the circumference, effectively making a circle with radius 7.

By graphing in this manner, you ensure that your circle is accurately represented on the coordinate plane, clearly showing its size and location. This not only helps in visual understanding but also in solving geometric problems involving circles.