Problem 22 - Calculus Early Transcendentals: Pearson New International Edition

Question

in Problems 17-22, find the center and radius of the circle with the given equation. \(x^{2}+16 x+\frac{105}{16}+4 y^{2}+3 y=0\)

Step-by-Step Solution

Answer

Center: (-8, -3/8), Radius: 5/8.

Step 1: Group the Terms

First, we will group the terms with the same variable together. The equation is given as:\[ x^{2} + 16x + \frac{105}{16} + 4y^{2} + 3y = 0. \]Let's rewrite it by grouping the \(x\) terms and the \(y\) terms separately:\[ (x^{2} + 16x) + (4y^{2} + 3y) + \frac{105}{16} = 0. \]

Step 2: Complete the Square for x

For the terms involving \(x\), complete the square:\(x^2 + 16x\).- The coefficient of \(x\) is 16. Half of this is 8, and squaring 8 gives 64.- Add and subtract 64 inside the equation to balance it:\[ (x^2 + 16x + 64 - 64). \]- This can be rewritten as:\[ (x+8)^2 - 64. \]

Step 3: Complete the Square for y

For the terms involving \(y\), complete the square by first factoring out 4:\(4y^2 + 3y\).- Factor out 4:\[ 4(y^2 + \frac{3}{4}y).\]- Take half of the coefficient for \(y\), which is \(\frac{3}{4}\), half of it is \(\frac{3}{8}\), and squaring it gives \(\frac{9}{64}\).- Add and subtract \(\frac{9}{64}\)) inside the grouped \(y\) terms:\[ 4(y^2 + \frac{3}{4}y + \frac{9}{64} - \frac{9}{64}) = 4((y + \frac{3}{8})^2 - \frac{9}{64}).\]

Step 4: Simplify the Equation

Replace both completed squares into the equation:\[(x + 8)^2 - 64 + 4\left((y + \frac{3}{8})^2 - \frac{9}{64}\right) + \frac{105}{16} = 0.\]Now multiply 4 by \(-\frac{9}{64}\):\[ (x + 8)^2 - 64 + 4(y + \frac{3}{8})^2 - \frac{36}{64} + \frac{105}{16} = 0. \]

Step 5: Solve for Radius and Center

Collect constants:- Convert \(-64\) to \(-\frac{1024}{16}\):\[-\frac{1024}{16} - \frac{36}{64} = -\frac{1024}{16} - \frac{9}{16} + \frac{105}{16} = -\frac{1024}{16} - \frac{9}{16} + \frac{105}{16}.\]Simplify:\[(x+8)^2 + 4(y + \frac{3}{8})^2 = \frac{25}{16}. \]This transforms to:\[(x+8)^2 + (y + \frac{3}{8})^2 = \frac{25}{64}. \]The center of the circle is \((-8, -\frac{3}{8})\) and the radius is \(\frac{5}{8}\).

Key Concepts

Completing the Square

Completing the square is a method used in algebra to transform quadratic equations into a form that makes them easier to solve or interpret. It involves creating a perfect square trinomial from a quadratic expression. This is crucial for rewriting equations of circles in a standard form, allowing us to identify vital properties of the circle, such as its center and radius.

• First, identify the quadratic terms to be transformed. In this example, we have quadratic terms in both variables: in \(x\) as \(x^2 + 16x\) and in \(y\) as \(4y^2 + 3y\).
• Next, for each variable, find the coefficient of the linear term, divide it by 2, square the result, and add and subtract this square within the respective expression. For \(x\), half of 16 is 8, and 8 squared is 64. Similarly, for \(y\), factor out the 4 first, to make it \(4(y^2 + \frac{3}{4}y)\), then proceed by taking half of \(\frac{3}{4}\), which is \(\frac{3}{8}\); square it to get \(\frac{9}{64}\).
• Adjust the equation to maintain equality by adding and subtracting the appropriate square values.

Through these steps, we systematically adjust the equation to reveal hidden structures like perfect square trinomials.

Center of a Circle

Finding the center of a circle from its equation involves understanding the standard form of a circle's equation: \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) is the center. After completing the square for both \(x\) and \(y\) terms in the given equation, we can rearrange it into the standard circle form.

• For our example, completing the square gives us the expression: \((x+8)^2 + (y + \frac{3}{8})^2 = \frac{25}{64}\).
• The circle's center \((h, k)\) corresponds to the terms inside the parentheses, with coordinates reversed in sign from those in the equation. Thus, \((h, k) = (-8, -\frac{3}{8})\).

This transformation is essential, as it provides a quick and clear way to visualize the circle's location on the coordinate plane.

Radius Calculation

The radius of a circle is computed from the standard form equation \((x-h)^2 + (y-k)^2 = r^2\), and specifically from the right-hand side, which represents the square of the radius \(r\).

• After completing the square and arranging the equation in the standard form \((x+8)^2 + (y+\frac{3}{8})^2 = \frac{25}{64}\), identify the term \(\frac{25}{64}\) on the right.
• The value \(\frac{25}{64}\) represents \(r^2\). The actual radius \(r\) is thus the square root of \(\frac{25}{64}\).
• Calculating this, \(r = \sqrt{\frac{25}{64}}\), which simplifies to \(\frac{5}{8}\).

The radius is a measure of how far each point of the circle is from its center, forming the basis of many geometric and analytical applications.