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Magazine Chapter 1: Problem 22
In Problems 17-22, find the center and radius of the circle with the given equation. x^{2}+16 x+\frac{105}{16}+4 y^{2}+3 y=0
Short Answer
Expert verified
The center is (-8, -3/8) and the radius is \( \sqrt{\frac{29}{8}} \).
Step by step solution
01
Rearrange the Equation
Start with the given equation: \( x^2 + 16x + \frac{105}{16} + 4y^2 + 3y = 0 \). Separate the \( x \) terms and \( y \) terms: \( x^2 + 16x + 4y^2 + 3y = -\frac{105}{16} \).
02
Complete the Square for x
To complete the square for \( x^2 + 16x \), take half of the coefficient of \( x \), square it, and add it to both sides. Half of 16 is 8, and 8 squared is 64. Add and subtract 64 inside the equation: \((x^2 + 16x + 64) - 64 \).
03
Complete the Square for y
For \( 4y^2 + 3y \), first factor out the 4: \( 4(y^2 + \frac{3}{4}y) \). Half of \( \frac{3}{4} \) is \( \frac{3}{8} \), and squaring it gives \( \frac{9}{64} \). Add and subtract \( \frac{9}{64} \) inside: \( 4((y + \frac{3}{8})^2 - \frac{9}{64}) \).
04
Simplify and Combine
Re-write the squares and constants: \( (x + 8)^2 - 64 + 4(y + \frac{3}{8})^2 - \frac{9}{16} = -\frac{105}{16} \). Combine constants: \( -64 = -\frac{1024}{16} \) and \( -\frac{9}{16} \) gives \( (x + 8)^2 + 4(y + \frac{3}{8})^2 = \frac{928}{16} \).
05
Solve for Circle Equation Standard Form
Convert this equation to standard circle form \( (x-h)^2 + (y-k)^2 = r^2 \) by isolating terms for \( y \) and dividing by 4: \( (x + 8)^2 + (y + \frac{3}{8})^2 = \frac{232}{16} \). Divide all terms by 4 to get \( \frac{232}{64} = \frac{29}{8} \).
06
Identify the Center and Radius
The center of the circle is \( (h, k) = (-8, -\frac{3}{8}) \) and the radius \( r \) is the square root of \( \frac{29}{8} \), which is \( \sqrt{\frac{29}{8}} \).
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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 in algebra to transform quadratic expressions into a perfect square trinomial. This method is particularly useful when dealing with equations of circles. Let's break it down a little more.
- For the expression involving x, we started with: \( x^2 + 16x \).
- We take half of 16, which is 8, then square it to get 64.
- Add and subtract 64 to form a complete square: \( (x + 8)^2 - 64 \).
- We have a factor of 4 out front, so isolate \( y^2 + \frac{3}{4}y \).
- Half of \( \frac{3}{4} \) is \( \frac{3}{8} \), and squaring it gives \( \frac{9}{64} \).
- Incorporate this to complete the square: \( 4((y + \frac{3}{8})^2 - \frac{9}{64}) \).
Circle Center
The circle center, represented as \((h, k)\), is derived from the equation's completed square form. In a circle equation of the standard form \((x-h)^2 + (y-k)^2 = r^2\), the center is given by the coordinates \((h, k)\).
For example, after completing the square for both x and y, the given equation takes the form \((x + 8)^2 + (y + \frac{3}{8})^2\). Here, compared to a standard circle equation, the coordinates of the center are extracted directly as \((h, k) = (-8, -\frac{3}{8})\). This step is key to understanding the circle's position on a coordinate plane.
For example, after completing the square for both x and y, the given equation takes the form \((x + 8)^2 + (y + \frac{3}{8})^2\). Here, compared to a standard circle equation, the coordinates of the center are extracted directly as \((h, k) = (-8, -\frac{3}{8})\). This step is key to understanding the circle's position on a coordinate plane.
Circle Radius
The radius of a circle is determined once the equation is transformed into its standard form. The standard equation of a circle is expressed as \((x-h)^2 + (y-k)^2 = r^2\). The variable \( r \) here represents the circle's radius, squared.
In our example, after simplifying and dividing terms accordingly, we find \((x + 8)^2 + (y + \frac{3}{8})^2 = \frac{29}{8}\).
To derive the radius, take the square root of the right-hand squared side:
In our example, after simplifying and dividing terms accordingly, we find \((x + 8)^2 + (y + \frac{3}{8})^2 = \frac{29}{8}\).
To derive the radius, take the square root of the right-hand squared side:
- \( r \) is the square root of \( \frac{29}{8} \), forming its actual length.
- This gives \( r = \sqrt{\frac{29}{8}} \).
Standard Form of a Circle
The standard form of a circle's equation is \((x-h)^2 + (y-k)^2 = r^2\). This equation is instrumental in coordinate geometry for identifying a circle's position. The transformation into standard form involves completing the square for both \( x \) and \( y \) terms.
- In the exercise, the equation was initially jumbled, and after systematic completion and constant adjustment, it formulated into \((x + 8)^2 + (y + \frac{3}{8})^2 = \frac{29}{8}\).
- From this form, both the circle's center, \((h,k) = (-8, -\frac{3}{8})\), and the radius \( r = \sqrt{\frac{29}{8}} \) are easily identified.
Coordinate Geometry
Coordinate geometry involves analyzing geometric shapes through their algebraic expression. The study of circles like our example relies heavily on this connection.
- The exercise demonstrates organizing components into a geometric entity on the plane.
- From organizing terms using the coordinate axes and completing the square, we extrapolate the necessary parameters: center and radius.
- Understanding it helps in transitioning between algebraic manipulation and visual representation.
