Problem 32 - Algebra A Combined Function

Question

The graph of each equation is a circle. Find the center and the radius, and then graph the circle. See Examples 5 through 7. $$ x^{2}+y^{2}-2 x-6 y-5=0 $$

Step-by-Step Solution

Answer

Center: (1, 3), Radius: \(\sqrt{15}\).

Step 1: Rearrange the Equation

First, rearrange the given equation to group the x and y terms together: \[ x^2 - 2x + y^2 - 6y = 5. \]

Step 2: Complete the Square for x Terms

To complete the square for x, take half of the coefficient of x which is -2, square it to get (-1)^2 = 1, and add it inside the equation: \[ x^2 - 2x + 1. \]

Step 3: Complete the Square for y Terms

Similarly, for the y terms, take half the coefficient of y, which is -6. Square it to get (-3)^2 = 9, and add it inside the equation: \[ y^2 - 6y + 9. \]

Step 4: Adjust the Equation for Additions

Because we added 1 and 9 inside the equation, we need to add these constants (1 + 9 = 10) to the right side as well: \[ x^2 - 2x + 1 + y^2 - 6y + 9 = 5 + 10. \]

Step 5: Simplify to Standard Circle Equation

Rewrite the left side as perfect squares and simplify the equation: \[ (x-1)^2 + (y-3)^2 = 15. \]

Step 6: identify the Center and Radius

The circle is now in the form \((x-h)^2 + (y-k)^2 = r^2\) which represents a circle with center \((h, k)\) and radius \(r\). Here \((h, k) = (1, 3)\) and \(r = \sqrt{15}\).

Key Concepts

Completing the Square

Completing the square is a mathematical technique used to simplify quadratic expressions. This method helps in transforming a quadratic equation into a perfect square trinomial, making it easier to solve problems related to circles, parabolas, and more. It involves a few steps that are easy to follow:

• Identify the quadratic expression, such as \[x^2 - 2x\] for the x terms.
• Take half of the linear coefficient (which is -2 here) and square it. Half of -2 is -1, and squaring it gives you \[(-1)^2 = 1\].
• Add this square (1) to the expression to form a perfect square trinomial: \[x^2 - 2x + 1\].
• Repeat the same process for the y terms, like \[y^2 - 6y\]. Half of -6 is -3, and its square is \[(-3)^2 = 9\]. Add this square to form \[y^2 - 6y + 9\].

These steps enable you to rewrite your quadratic expression into a perfect square form, making the equation more manageable and eventually leading you to the circle's standard equation.

Center of a Circle

The center of a circle in coordinate geometry is a crucial point, represented as \((h, k)\). It signifies the circle's central position on the plane. When you have an equation structured in the standard circle form \((x-h)^2 + (y-k)^2 = r^2\), the center can be directly identified.

• Start by completing the square for the x and y terms in a given equation. This restructuring leads to a simple comparison with the standard form.
• In our example, after completing the square, we reached \((x-1)^2 + (y-3)^2 = 15\).
• Here, the values of \(h = 1\) and \(k = 3\) indicate that the circle's center is at point \((1, 3)\).

Understanding the circle’s center is vital for graphing and analyzing the circle's position and relation with other geometric figures on the coordinate plane.

Radius of a Circle

The radius of a circle is the distance from the circle's center to any point on its boundary. It is crucial for understanding the size and space a circle occupies. In the standard circle equation \((x-h)^2 + (y-k)^2 = r^2\), the term \(r^2\) represents the square of the radius.

• From the completed equation, consider \((x-1)^2 + (y-3)^2 = 15\).
• Here, \(15\) stands for \(r^2\), therefore the radius \(r = \sqrt{15}\).
• So, the circle’s radius is \(\sqrt{15}\), which is an irrational number, approximately equal to 3.87.

Calculating the radius helps in understanding how large a circle is and is useful for various applications, such as when calculating the circumference or the area of the circle.