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Chapter 2: Problem 79

How is the standard form of a circle鈥檚 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 grouping \(x\) and \(y\) terms, factoring out the coefficients if not 1, completing the square for each grouped term, and comparing the resulting equation to the standard form to find the centre and radius of the circle.

Step by step solution

01

Factor Out Coefficients

Begin by grouping the \(x\) and \(y\) terms. If the coefficients of \(x^2\) and \(y^2\) are not 1, factor them out: \(A(x^2 + \frac{D}{A}x) + C(y^2 + \frac{E}{C}y) + F = 0\)
02

Complete the Square

To complete the square for each grouped term, add the square of half the coefficient of the linear term. Ensure to maintain equality by adding these terms to both sides of the equation: \[A(x^2 + \frac{D}{A}x + (\frac{D}{2A})^2) + C(y^2 + \frac{E}{C}y + (\frac{E}{2C})^2) + F' = 0\] \(F' = F + A(\frac{D}{2A})^2 + C(\frac{E}{2C})^2\) where \(F'\) is the result of modifications made to preserve the balance of the equation.
03

Rewrite as Square of Binomials

Now, re-write the \(x\) and \(y\) grouped terms as the square of binomials, and the right side of the equation as the square of radius: \[(x + \frac{D}{2A})^2 + (y + \frac{E}{2C})^2 = \frac{F'}{A + C}\] Compare this equation with the standard form to obtain the values of \((a, b)\) and \(r^2\).
04

Obtaining the Centre and Radius

The centre of the circle \((a,b)\) and the radius \(r\) can be found by comparing the derived equation to the standard form. The centre of the circle is at \(-\frac{D}{2A}\), \(-\frac{E}{2C}\) and the radius is equal to the square root of \(\frac{F'}{A + C}\).

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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 method used to transform a quadratic expression into a perfect square trinomial. This process is essential when working with circle equations in general form. It helps to convert the equation into the standard form, making it easier to identify the center and radius of the circle.

To complete the square:
  • Group the quadratic and linear terms separately for both \(x\) and \(y\).
  • Find the square of half the coefficient of the linear term. This value is what we add and subtract to complete the square.
  • Add this term inside the equation while maintaining balance by adjusting the constant on the other side.
By using these steps, you make the expression easier to manipulate and understand, especially when you transform it into the standard form of a circle's equation.
Standard Form of a Circle
The standard form of a circle鈥檚 equation makes it straightforward to identify the circle's properties. It is expressed as:\[(x - a)^2 + (y - b)^2 = r^2\]where \( (a, b) \) is the center of the circle and \(r\) is the radius.

Transforming a circle's equation from its general form to this standard form involves:
  • Completing the square for both \(x\) and \(y\) terms.
  • Ensuring the equation's balance by adjusting constants appropriately.
Once in standard form, it readily provides insights into the geometric properties of the circle, such as its position in the coordinate plane and its size.
Radius of a Circle
The radius of a circle is one of the critical components derived from its equation in standard form. It represents the distance from the center of the circle to any point on its circumference. Once the circle's equation is in the format \[(x - a)^2 + (y - b)^2 = r^2\],extracting the radius is straightforward.

The radius \(r\) is found by taking the square root of the constant on the right-hand side of the equation. This means:
  • In the equation, \(r = \sqrt{r^2}\).
  • Ensure \(r\) is always a positive value, representing a true geometric distance.
This simplicity makes the radius a convenient measure for understanding the size and proportion of the circle in its standard form.
Center of a Circle
The center of a circle is an essential aspect, giving it a defined position in the coordinate plane. In the standard form of a circle's equation, the center coordinates are easily identifiable. The equation \[(x - a)^2 + (y - b)^2 = r^2\]highlights that the circle's center is located at the point \( (a, b) \).

To determine the center when converting from the general form:
  • First, complete the square for the \(x\) and \(y\) terms of the equation.
  • The values derived from these squares, when negated, are the \(a\) and \(b\) coordinates of the center.
Understanding the center's location is crucial for graphing the circle and analyzing its position relative to other geometric figures and axes. It provides essential spatial context for further mathematical exploration and problem-solving.

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Most popular questions from this chapter

Use a graphing utility to graph each equation in Exercises \(100-103\). Then use the \([\text { TRACE }]\) feature to trace along the line and find the coordinates of two points. Use these points to compute the line's slope. Check your result by using the coefficient of \(x\) in the line's equation. $$ y=2 x+4 $$

In your own words, describe how to find the distance between two points in the rectangular coordinate system.

find and simplify the difference quotient $$\frac{f(x+h)-f(x)}{h}, h \neq 0$$ for the given function. $$ f(x)=\frac{1}{x} $$

Express the given function \(h\) as \(a\) composition of two functions \(f\) and \(g\) so that \(h(x)=(f \circ g)(x)\). $$h(x)=\frac{1}{4 x+5}$$

Let \(f\) and \(g\) be defined by the following table: \(\begin{array}{rrr}{x} & {f(x)} & {g(x)} \\ {-2} & {6} & {0} \\ {-1} & {3} & {4} \\ {0} & {-1} & {1} \\ {1} & {-4} & {-3} \\ {2} & {0} & {-6}\end{array}\) Find \(\sqrt{f(-1)-f(0)}-[g(2)]^{2}+f(-2) \div g(2) \cdot g(-1)\)
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