Chapter 13: Problem 28
The graph of each equation is a circle. Find the center and the radius, and then graph the circle. See Examples 5 through 7. $$ x^{2}+y^{2}+6 x+10 y-2=0 $$
Short Answer
Expert verified
The center is (-3, -5) and the radius is 6.
Step by step solution
01
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.
02
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:\[ x^2 + 6x = (x + 3)^2 - 9 \].
03
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:\[ y^2 + 10y = (y + 5)^2 - 25 \].
04
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:\[(x+3)^2 + (y+5)^2 = 36\].
05
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 \( r = \sqrt{36} = 6 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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 .
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).
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 \(\sqrt{36} = 6\).