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Chapter 2: Problem 54

Exer. \(47-56:\) Find the center and radius of the circle with the given equation. $$x^{2}+y^{2}-6 x+4 y+13=0$$

Short Answer

Expert verified
Center: (3, -2), Radius: 0

Step by step solution

01

Rewrite the Circle Equation

First, we need to rewrite the given equation in a form that makes it easier to find the center and radius. The equation is currently: \(x^{2}+y^{2}-6x+4y+13=0\). Our goal is to write it in the standard form: \((x-h)^{2}+(y-k)^{2}=r^{2}\).
02

Rearrange the Equation

Move the constant term to the other side of the equation to prepare for completing the square. This gives us: \(x^{2} - 6x + y^{2} + 4y = -13\).
03

Complete the Square for x

To complete the square for the x-terms, take the coefficient of x, divide it by 2, and square it. The coefficient is \(-6\), so \((-6/2)^{2} = 9\). Add and subtract 9 inside the x-terms: \(x^{2} - 6x + 9\). This becomes \((x-3)^{2}\). Add 9 to the right side as well: \(-13 + 9 = -4\).
04

Complete the Square for y

For the y-terms, take the coefficient of y, divide it by 2, and square it. The coefficient is \(4\), so \((4/2)^{2} = 4\). Add and subtract 4 inside the y-terms: \(y^{2} + 4y + 4\). This becomes \((y+2)^{2}\). Add 4 to the right side as well: \(-4 + 4 = 0\).
05

Form the Standard Circle Equation

Combine the completed square terms to form the equation \((x-3)^{2} + (y+2)^{2} = 0\). Now the equation is in the standard form for a circle, where \(h=3\), \(k=-2\), and \(r^{2}=0\), meaning \(r=0\).
06

Identify the Center and Radius

From the standard form \((x-3)^{2} + (y+2)^{2} = 0\), we identify the center of the circle as \((3, -2)\) and the radius as \(r = 0\).

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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 technique used to transform a quadratic equation into a perfect square trinomial. This method simplifies the equation to reveal hidden structures. Here's the breakdown:
  • Identify the quadratic term, the coefficient of the linear term, and the constant term.
  • Take the linear term's coefficient, divide it by 2, and square the result. This number is called the "completing square" element.
  • Add and subtract this number within the equation to transform it into a perfect square trinomial.
In our original equation, for the x term, the coefficient is -6. Dividing by 2 gives us -3, and squaring -3 results in 9. Similarly, for the y term, the coefficient is 4. Splitting it gives 2, and squaring gives 4. By adding and adjusting these values, we can turn the equation into a form that is easier to understand and analyze.
Standard Form of a Circle
The standard form of a circle's equation is pivotal for finding circle properties like the center and radius. This form is written as:\[(x-h)^2 + (y-k)^2 = r^2\]Here,
  • \( (h, k) \) represents the center of the circle.
  • \( r \) is the radius of the circle.
To rearrange a given circle equation to this form, we employ completing the square for both x and y terms, as shown in the solution. When a circle's equation is expressed in this standard format, identifying its core parameters becomes a simple task.
This reformatting helps us clearly see the operations required to get to the derived formula, providing better insights into the geometric properties of the circle.
Center of a Circle
The center of a circle is a crucial point located at the heart of the circle's geometry. It is described by the coordinates \( (h, k) \) in the standard circle equation \( (x-h)^2 + (y-k)^2 = r^2\).
  • "h" is the x-coordinate of the center.
  • "k" is the y-coordinate of the center.
Once the equation is in standard form, identifying the center is straightforward. In the provided exercise, the equation was approached to reveal it as \((x-3)^2 + (y+2)^2 = 0\). Clearly, the center of the circle is at \( (3, -2) \).
The center gives us a fixed point around which every point on the circle maintains a constant distance, which leads us to the concept of the radius.
Radius of a Circle
The radius is another fundamental component of a circle, representing the distance from the center to any point on the circle's circumference. In its standard form, a circle's equation can be expressed as \( (x-h)^2 + (y-k)^2 = r^2 \).
  • The term \( r^2 \) corresponds directly to the square of the radius.
  • To find the radius, take the square root of \( r^2 \).
In the provided solution, the circle's radius was calculated by noting that \( r^2 = 0 \), implying a radius \( r = 0 \). This particular circle is just a point located at its center since a zero radius means there is no extension from the center point outward.

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Most popular questions from this chapter

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