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Chapter 1: Problem 118

Find the center and radius of the circle \(x^{2}+3 x+y^{2}-4 y=0\)

Short Answer

Expert verified
The center is \((-\frac{3}{2}, 2)\) and the radius is \(\frac{5}{2}\).

Step by step solution

01

Group Terms

First, group the x terms and the y terms together: \[ x^2 + 3x + y^2 - 4y = 0 \]
02

Complete the Square for x Terms

To complete the square for the x terms, take half of the coefficient of x, square it, and add/subtract it within the equation: \( x^2 + 3x = (x^2 + 3x + (\frac{3}{2})^2) - (\frac{3}{2})^2 = (x + \frac{3}{2})^2 - \frac{9}{4} \).
03

Complete the Square for y Terms

Do the same for the y terms: \( y^2 - 4y = (y^2 - 4y + (-2)^2) - (-2)^2 = (y - 2)^2 - 4 \).
04

Combine and Simplify

Combine the completed squares and move the constants to the other side of the equation: \[ (x + \frac{3}{2})^2 - \frac{9}{4} + (y - 2)^2 - 4 = 0 \]\[ (x + \frac{3}{2})^2 + (y - 2)^2 = \frac{25}{4} \]
05

Identify Center and Radius

Recognize the standard form of the circle equation \[ (x - h)^2 + (y - k)^2 = r^2 \]From this, the center (h, k) = (-\frac{3}{2}, 2) \) and the radius \( r = \frac{5}{2} \.

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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 method used to convert a quadratic equation into a perfect square trinomial. This is particularly useful when finding the center and radius of a circle from its general equation.
To complete the square for the term involving x in the equation \( x^2 + 3x \), follow these steps:
  • Take half of the coefficient of x, which is 3. So, half of 3 is \( \frac{3}{2} \).
  • Square \( \frac{3}{2} \) to get \( ( \frac{3}{2} )^2 = \frac{9}{4} \).
  • Add and subtract \( \frac{9}{4} \) within the equation to balance it. This step transforms \( x^2 + 3x \) into \( (x + \frac{3}{2})^2 - \frac{9}{4} \).
Repeat the process for the y terms. The initial term \( y^2 - 4y \) involves:
  • Half of -4 is -2.
  • Square -2, resulting in 4.
  • Add and subtract 4 in the equation. The term then changes to \( (y - 2)^2 - 4 \).
Having completed the square for both x and y terms, we transition the equation into its standard form.
circle equation
The general form of a circle鈥檚 equation is \( Ax^2 + Ay^2 + Bx + Cy + D = 0 \). To identify the circle's center and radius, we need the standard form, \( (x - h)^2 + (y - k)^2 = r^2 \).
Using the given exercise, our starting point was:
\( x^2 + 3x + y^2 - 4y = 0 \). By grouping and completing the square, we transformed this into:
\( (x + \frac{3}{2})^2 - \frac{9}{4} + (y - 2)^2 - 4 = 0 \).
Then, combining terms and isolating the sum of squares on one side, we find:
\( (x + \frac{3}{2})^2 + (y - 2)^2 = \frac{25}{4} \).
This new form conveniently mirrors the standard circle equation, making it easy to pinpoint the center \( (h, k) \) and radius \( r \).
standard form of a circle
The standard form of a circle鈥檚 equation simplifies identifying the circle's center and radius. It is expressed as:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Here, \( (h, k) \) represents the center coordinates, and \( r \) is the radius.
For our example, the transformed equation became:
\[ (x + \frac{3}{2})^2 + (y - 2)^2 = \frac{25}{4} \]
Identifying the standard form components:
  • Center \( (h, k) = (- \frac{3}{2}, 2) \)
  • Radius \( r = \frac{5}{2} \)
Understanding these components is crucial. The terms within the parentheses define the horizontal and vertical shifts from the origin, effectively centering the circle. The right side of the equation represents the squared radius, hence giving the actual radius upon taking its square root.

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