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Chapter 11: Problem 108

Determine the center and radius of the circle \(x^{2}-10 x+y^{2}+2 y=0\)

Short Answer

Expert verified
The center is (5, -1) and the radius is \sqrt{26}.

Step by step solution

01

- Rewrite the Equation in Standard Circle Form

The given equation is: \[ x^{2} - 10x + y^{2} + 2y = 0 \]. To find the center and radius, rewrite this equation in the standard form of a circle equation: \[ (x-h)^{2} + (y-k)^{2} = r^{2} \].
02

- Complete the Square for the x Terms

Take the x terms \[ x^{2} - 10x \]. To complete the square, add and subtract \[ \frac{(-10)^2}{4} = 25 \]: \[ x^{2} - 10x + 25 - 25 \]. This can be rewritten as: \[ (x - 5)^{2} - 25 \].
03

- Complete the Square for the y Terms

Now take the y terms \[ y^{2} + 2y \]. To complete the square, add and subtract \[ \frac{2^2}{4} = 1 \]: \[ y^{2} + 2y + 1 - 1 \]. This can be rewritten as: \[ (y + 1)^{2} - 1 \].
04

- Rewrite the Equation with Completed Squares

Combine the results from Steps 2 and 3 into the original equation: \[ (x - 5)^{2} - 25 + (y + 1)^{2} - 1 = 0 \].
05

- Simplify the Equation

Simplify the equation by moving constants to the other side: \[ (x - 5)^{2} + (y + 1)^{2} - 25 - 1 = 0 \] \[ (x - 5)^{2} + (y + 1)^{2} = 26 \]. This is now in the form \[ (x-h)^{2} + (y-k)^{2} = r^{2} \].
06

- Identify the Center and Radius

From the simplified equation \[ (x - 5)^{2} + (y + 1)^{2} = 26 \], identify the center \[(h, k) = (5, -1)\] and the radius \( r = \sqrt{26} \).

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Key Concepts

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

center of a circle
To find the center of a circle from its equation, you need to put the equation into its standard form. This form is: ewline (x-h)^2 + (y-k)^2 = r^2 ewline where (h,k) represents the center of the circle. ewline Steps to identify the center: ewline
  • Identify the x-terms and complete the square.
  • Identify the y-terms and complete the square.
  • Combine these completed squares into one equation.
  • The values of h and k from the resulting equation will give you the coordinates of the center of the circle.
For example, from the given equation ewline x^2 - 10x + y^2 + 2y = 0 ewline after completing the square, we rewrite it as ewline (x-5)^2 + (y+1)^2 = 26. ewline Here, (h,k) = (5,-1), is the center of the circle.
radius of a circle
The radius of a circle in the standard equation form ewline (x-h)^2 + (y-k)^2 = r^2 ewline is represented by r. ewline To find the radius:
  • First, complete the square for both x and y terms to get the equation in standard form.
  • Identify the value on the right-hand side of the equation, which is r^2.
  • Take the square root of this value to get the radius, r.
From the example equation: ewline (x-5)^2 + (y+1)^2 = 26, ewline the value on the right-hand side is 26, so r^2 = 26. ewline Taking the square root of 26, we get the radius r = sqrt(26).
completing the square
Completing the square is a method used to transform quadratic equations into a form that allows easy identification of key properties. ewline Steps to complete the square:
  • For the x terms, take the coefficient of x, divide it by 2, and square it.
  • Add and subtract this squared value within the equation.
  • Rewrite the expression as a squared binomial.
  • Repeat the same process for the y terms.
Let's see how this works with x^2 - 10x:
  • Identify the coefficient of x, which is -10.
  • Divide by 2 to get -5, then square it to get 25.
  • Add and subtract 25 inside the equation: x^2 - 10x + 25 - 25
  • Rewrite as (x-5)^2 - 25
Similarly, for y^2 + 2y:
  • Identify 2, divide by 2 to get 1, then square it to get 1.
  • Add and subtract 1: y^2 + 2y + 1 - 1
  • Rewrite as (y+1)^2 - 1

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