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Chapter 1: Problem 61

Find the center and the radius of the circle given by the equation \(0=x^{2}+y^{2}+6 x+2 y-12\). (To do this, you must complete the squares.)

Short Answer

Expert verified
Center: (-3, -1); Radius: \(\sqrt{22}\).

Step by step solution

01

Write the Equation with Like Terms Together

Start by rewriting the equation to group the x and y terms together. The given equation is: \[0 = x^2 + 6x + y^2 + 2y - 12\] Re-order it to focus on the x's and y's:\[x^2 + 6x + y^2 + 2y = 12\]
02

Complete the Square for the x Terms

For the terms involving x, which are \(x^2 + 6x\), we complete the square. Take half of the coefficient of x (which is 6), square it, and then add and subtract that square:\[x^2 + 6x = (x^2 + 6x + 9) - 9 = (x + 3)^2 - 9\]
03

Complete the Square for the y Terms

Next, complete the square for the terms involving y: \(y^2 + 2y\). Take half of the coefficient of y (which is 2), square it, then add and subtract that square:\[y^2 + 2y = (y^2 + 2y + 1) - 1 = (y + 1)^2 - 1\]
04

Rewrite the Equation with Completed Squares

Substitute the completed squares back into the main equation:\[(x + 3)^2 - 9 + (y + 1)^2 - 1 = 12\]Combine like terms:\[(x + 3)^2 + (y + 1)^2 = 12 + 9 + 1 = 22\]
05

Identify the Center and Radius

The circle equation is now in standard form:\[(x - (-3))^2 + (y - (-1))^2 = 22\]The center \((h, k) = (-3, -1)\) and the radius \(r = \sqrt{22}\).

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Key Concepts

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

Circle Equation
The circle equation is a special way to express all the points \((x, y)\) that lie on a circle. It's structured as follows:
  • The left side includes terms for both \(x\) and \(y\), typically squared.
  • The right side generally represents a constant value, showing how far each point is from the circle's center.
In the exercise given, we started with the equation \[0 = x^2 + y^2 + 6x + 2y - 12\] This equation does not look exactly like our typical circle equation at first glance. So, we must adjust it using algebraic techniques like completing the square, to express it in a recognizable form. The goal is to identify the circle's characteristics such as its center and radius.
Center and Radius
The center and radius are crucial for describing a circle's position and size. When the circle equation is in the proper form, \((x - h)^2 + (y - k)^2 = r^2\), these features are easy to find.

  • The center \((h, k)\) is determined directly from the equation's structure. Here, \((x - h)\) and \((y - k)\) inform us of how the equation is offset from origin points (0, 0).
  • The radius \(r\) is simply the square root of the constant term on the right side of the equation.
In our specific problem, after completing the squares, we found the center at \((-3, -1)\). The radius is determined as \(\sqrt{22}\). Knowing these allows you to visualize where the circle is on a grid, and how far it extends.
Standard Form of a Circle
To convey the geometry of a circle, the standard form of a circle's equation is very handy: \[(x - h)^2 + (y - k)^2 = r^2\]

  • This form makes it straightforward to read off the circle's center \( (h, k) \) and radius \( r \).
  • It simplifies identifying and working with the circle's properties.
In our exercise, the given equation was not initially in this standard form. Through completing the square method, we reshaped the equation to \[(x + 3)^2 + (y + 1)^2 = 22\]. This expression is easy to interpret:
  • The center is at \((-3, -1)\)
  • The radius is \(\sqrt{22}\)
Working in the standard form provides clarity, allowing you to easily interpret and work with circles in the coordinate plane.

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