Problem 28 - 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. $$ x^{2}+y^{2}+6 x+10 y-2=0 $$

Step-by-Step Solution

Answer

The center is (-3, -5) and the radius is 6.

Step 1: Rewrite the Equation

First, rewrite the given circle equation in the form \[ x^2 + y^2 + 6x + 10y - 2 = 0 \]by separating the terms involving x and y.

Step 2: Complete the Square for x

To complete the square for terms involving x, take the terms \( x^2 + 6x \). Add and subtract \( (6/2)^2 = 9 \) to form a perfect square:\[ x^2 + 6x = (x + 3)^2 - 9 \].

Step 3: Complete the Square for y

To complete the square for terms involving y, take the terms \( y^2 + 10y \). Add and subtract \( (10/2)^2 = 25 \) to form a perfect square:\[ y^2 + 10y = (y + 5)^2 - 25 \].

Step 4: Simplify the Equation

Now substitute the completed squares back into the original equation:\[(x + 3)^2 - 9 + (y + 5)^2 - 25 - 2 = 0\]Combine and simplify the constants:\[(x+3)^2 + (y+5)^2 = 36\].

Step 5: Identify the Center and Radius

The standard form of a circle is \((x-h)^2 + (y-k)^2 = r^2\) where \((h, k)\) is the center and \(r\) is the radius. From \((x + 3)^2 + (y + 5)^2 = 36\), compare:- Center is \((-3, -5)\).- Radius \( r = \sqrt{36} = 6 \).

Key Concepts

Completing the Square

Completing the square is a mathematical technique used to transform a quadratic equation into a perfect square trinomial. It simplifies finding the core properties of a circle when given its general equation. Here's how it applies specifically to the circle equation:

• First, we focus on the quadratic terms independently within the equation. Taking terms separately for x ( x^2 + 6x ) and for y ( y^2 + 10y ).
• For each, add and subtract a calculated value to form perfect squares. For example, with x, you calculate the square of half the linear coefficient, i.e., (6/2)^2 = 9 .
• This turns x^2 + 6x into (x + 3)^2 - 9 ; it completes the square for terms involving x.
• Repeat similarly for y, taking (10/2)^2 = 25 to transform y^2 + 10y into (y + 5)^2 - 25 .

Completing the square allows us to rewrite the original circle equation in a standard form, facilitating easier identification of its center and radius. It systematically breaks down complex quadratic expressions to help visualize the geometric properties of circles.

Center of a Circle

When talking about a circle's equation, identifying the center is crucial. In the standard form of a circle's equation \((x-h)^2 + (y-k)^2 = r^2\), (h, k) represents the center of the circle.

• In our completed equation \((x + 3)^2 + (y + 5)^2 = 36\), compare and find the center: by matching this with the standard form.
• Observe that \(h\) gets a value of -3, and \(k\) is -5; hence, the center is at point (-3, -5).

Knowing the center of a circle is essential for graphing and understanding its position in coordinate geometry. It helps in plotting the circle correctly on the graph and visualizes how the circle is placed relative to other shapes and axes.

Radius of a Circle

The radius is another key feature of a circle. From the standard circle equation \((x-h)^2 + (y-k)^2 = r^2\), the radius \(r\) is determined by the constant on the right side of the equation.

• In the example, we simplify the equation to \((x+3)^2 + (y+5)^2 = 36\). This can be directly compared to the standard form.
• The constant 36 on the right side is equal to \(r^2\); thus, the radius \(r\) of the circle is calculated as \(\sqrt{36} = 6\).

The radius informs us about the size of the circle, allowing us to draw it to scale when graphing. It represents the fixed distance from the circle's center to any point on its circumference. Understanding both the radius and center is key to any calculations involving circles.