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Chapter 1: Problem 79

How is the standard form of a circle's equation obtained from its general form?

Short Answer

Expert verified
The standard form of a circle's equation is obtained from its general form by dividing through by the coefficient of the quadratic terms, moving the constant to the right side, completing the square on the left side and rearranging the equation to its standard form.

Step by step solution

01

Make the quadratic terms have a coefficient of 1

From the general form, make the coefficient of the quadratic terms to be 1 by dividing the entire equation through by A (assuming A=C and not zero). The equation becomes \(x^2 + y^2 + (D/A)x + (E/A)y + F/A = 0\).
02

Move the constant to the right

Move the constant term to the right side of the equation resulting in \(x^2 + y^2 + (D/A)x + (E/A)y = - F/A\).
03

Complete the square

Complete the squares on the left side of the equation by adding \(((D/ 2A) ^2)\) and \(((E/ 2A) ^2)\) to both sides. This gives \(x^2 + (D/A)x + (D/2A)^2 + y^2 + (E/A)y + (E/2A)^2 = (D/2A)^2 + (E/2A)^2 - F/A\).
04

Rearrange to Standard Form

Finally, rewrite the left side as two squared binomials and the right side as a squared radius. The realized equation is \((x + D/2A)^2 + (y + E/2A)^2 = r^2\), where \(r^2 = (D/2A)^2 + (E/2A)^2 - F/A\) which is now in standard form.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

General Form of a Circle
The general form of a circle's equation expresses the circle in terms of a quadratic equation for both the x and y variables. It appears as: \( Ax^2 + Ay^2 + Dx + Ey + F = 0 \). This equation is quite broad and can represent a variety of conic sections when A and C vary. However, when the coefficients of \(x^2\) and \(y^2\) are equal, it specifically represents a circle. The general form is advantageous because of its straightforward expansion and ease of comparison with other equations. To convert this into a more recognizable form, certain steps must be followed to simplify it and reveal the circle's essential properties, such as its center and radius.
Standard Form of a Circle
The standard form of a circle provides a clear and concise representation of a circle, making it easy to identify the circle's center and radius at a glance. This form is written as: \( (x - h)^2 + (y - k)^2 = r^2 \), where \( (h, k) \) represents the coordinates of the circle鈥檚 center, and \( r \) is the radius. Transitioning from the general form to this form involves a specific process, often requiring you to complete the square for both x and y terms. The standard form is particularly useful in graphing because it immediately gives insight into the circle's location and size, aiding in visual representation.
Completing the Square
Completing the square is a vital algebraic process used to convert the general form of a circle's equation to the standard form. Here's how it works: starting with the terms \( x^2 + (D/A)x \), you transform them into a perfect square trinomial. This involves adding and subtracting \((D/2A)^2 \). The same process is applied to the y terms \( y^2 + (E/A)y \), where you add and subtract \((E/2A)^2 \). Doing this reorganizes the equation so that the x and y components form squared binomials: \( (x + D/2A)^2 \) and \( (y + E/2A)^2 \). These steps reveal the center and radius of the circle, encapsulating all components within the elegant standard form. Completing the square thus simplifies mathematical expressions, allowing greater clarity in the circle's graphical and algebraic analyses.

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Most popular questions from this chapter

Determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. If \(f(x)=\sqrt{x}\) and \(g(x)=2 x-1,\) then \((f \circ g)(5)=g(2)\)

Explain how the vertical line test is used to determine whether a graph represents a function.

Determine whether each statement makes sense or does not make sense, and explain your reasoning. I must have made a mistake in finding the composite functions \(f \circ g\) and \(g \circ f,\) because I notice that \(f \circ g\) is not the same function as \(g \circ f\)

Determine whether each statement makes sense or does not make sense, and explain your reasoning. I used a function to model data from 1990 through 2015 . The independent variable in my model represented the number of years after \(1990,\) so the function's domain was \(\\{x | x=0,1,2,3, \dots, 25\\}\)

In Exercises \(53-64,\) complete the square and write the equation in standard form. Then give the center and radius of each circle and graph the equation. $$x^{2}+y^{2}+6 x+2 y+6=0$$
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