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Magazine Chapter 10: Problem 57
Find the center and radius of each circle. $$x^{2}+y^{2}+2 y=8$$
Short Answer
Expert verified
The center is (0, -1) and the radius is 3.
Step by step solution
01
Rewrite the equation in standard form
The given equation is \[ x^2 + y^2 + 2y = 8 \] We need to rewrite it in the standard form of a circle equation: \[ (x - h)^2 + (y - k)^2 = r^2 \].
02
Move constant term to the right side
Begin by isolating the constant term on the right side of the equation: \[ x^2 + y^2 + 2y - 8 = 0 \] Add 8 to both sides of the equation to isolate the constant term: \[ x^2 + y^2 + 2y = 8 \]
03
Complete the square for the y-terms
To complete the square for the y-terms, we need to add and subtract a constant inside the equation to form a perfect square trinomial. Observe the y-term 2y: Take half of the coefficient of y, which is 2, yielding 1, and square it: Half of 2 is 1, and \(1^2 = 1\). Add and subtract this square inside the equation: \[ x^2 + y^2 + 2y + 1 - 1 = 8 \] Rewrite the equation in a perfect square form: \[ x^2 + (y + 1)^2 - 1 = 8 \]
04
Simplify the equation
Combine the constants on the right side to get the standard form: \[ x^2 + (y + 1)^2 = 9 \] Now we have the equation in the standard form \[ (x - h)^2 + (y - k)^2 = r^2 \].
05
Identify the center and the radius
Compare \[ x^2 + (y + 1)^2 = 9 \] with \[ (x - h)^2 + (y - k)^2 = r^2 \]. Here, the center \((h, k)\) is \[(h, k) = (0, -1)\] The radius r is \(r = \sqrt{9} = 3\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
standard form of a circle
The standard form of a circle equation is a crucial concept in precalculus. It helps us identify and work with the fundamental elements of a circle, such as its center and radius. The formula for the standard form is given by:
(x - h)^2 + (y - k)^2 = r^2
Here,
(x - h)^2 + (y - k)^2 = r^2
Here,
- (h, k) represents the center of the circle.
- r represents the radius of the circle.
completing the square
Completing the square is an essential algebraic technique used to transform a quadratic equation into a perfect square trinomial. This is particularly useful when dealing with circle equations because it helps to convert them into the standard form. Here鈥檚 a general method to complete the square:
In our example, to complete the square for y^2 + 2y, we take half of 2, which is 1, and then square it to get 1. We then add and subtract this value:
y^2 + 2y + 1 - 1
This becomes (y + 1)^2 - 1, which simplifies our equation considerably.
- Identify the quadratic term and the linear term (e.g., in the expression x^2 + bx, identify x^2 and bx).
- Take half of the coefficient of the linear term, then square it.
- Add and subtract this square within the equation to form a perfect square trinomial. For example, with x^2 + bx, half of b squared is (b/2)^2. Add and subtract (b/2)^2.
- Rewrite the expression as a square of a binomial.
In our example, to complete the square for y^2 + 2y, we take half of 2, which is 1, and then square it to get 1. We then add and subtract this value:
y^2 + 2y + 1 - 1
This becomes (y + 1)^2 - 1, which simplifies our equation considerably.
center and radius of a circle
Knowing how to find the center and radius of a circle is fundamental in solving circle-related problems in precalculus. From our previous conversion into standard form, we can extract these values easily.
The general standard form equation is:
(x - h)^2 + (y - k)^2 = r^2
Comparing this with our equation x^2 + (y + 1)^2 = 9, we identify:
Thus, the center is (0, -1) and the radius is 3. These coordinates and measurements define the size and location of the circle on the plane.
The general standard form equation is:
(x - h)^2 + (y - k)^2 = r^2
Comparing this with our equation x^2 + (y + 1)^2 = 9, we identify:
- The center (h, k) = (0, -1), since there's no horizontal shift.
- The radius r can be found by taking the square root of 9, which is 3.
Thus, the center is (0, -1) and the radius is 3. These coordinates and measurements define the size and location of the circle on the plane.
perfect square trinomial
A perfect square trinomial is a quadratic expression that can be factored into a square of a binomial. It takes the form of:
(ax + b)^2 or (ax - b)^2
It results from the process of completing the square. For instance, in our example:
y^2 + 2y + 1
Here鈥檚 how it fits the structure of a perfect square trinomial:
(ax + b)^2 or (ax - b)^2
It results from the process of completing the square. For instance, in our example:
y^2 + 2y + 1
Here鈥檚 how it fits the structure of a perfect square trinomial:
- First term: y^2 (which is y squared)
- Middle term: 2y (which is twice the product of y and 1)
- Last term: 1 (which is the square of 1)
precalculus problems
Understanding circle equations and their transformations is a common topic in precalculus. Mastery of these concepts allows students to handle more complex problems with ease. Problems may involve:
To tackle such problems effectively, one should practice:
By consistently practicing these skills, students will find that precalculus circle problems become much more manageable and intuitive.
- Converting general form equations to standard form
- Identifying the center and radius from given equations
- Graphing circles on the coordinate plane
- Solving real-world problems involving circular shapes, like finding areas and lengths
To tackle such problems effectively, one should practice:
- Rewriting equations in standard form by completing the square
- Recognizing and forming perfect square trinomials
- Visualizing geometric transformations
By consistently practicing these skills, students will find that precalculus circle problems become much more manageable and intuitive.
