Problem 41 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the distance between the points on the circle \(x^{2}+2 x+y^{2}-2 y=20\) with the \(x\)-coordinates \(-2\) and 2 . How many such distances are there?

Step-by-Step Solution

Answer

There are 4 possible distances between the points.

Step 1: Understand the Circle Equation

The given equation of the circle is \( x^2 + 2x + y^2 - 2y = 20 \). We need to rewrite it in the standard form by completing the square for both \(x\) and \(y\).

Step 2: Complete the Square for x

For the term \( x^2 + 2x \), complete the square by adding and subtracting \(1\) inside the equation. This gives us:\[ x^2 + 2x = (x+1)^2 - 1 \]

Step 3: Complete the Square for y

For the term \( y^2 - 2y \), complete the square by adding and subtracting \(1\) inside the equation. This gives us:\[ y^2 - 2y = (y-1)^2 - 1 \]

Step 4: Rewrite the Circle Equation

Substitute both completed squares back into the equation:\[ (x+1)^2 - 1 + (y-1)^2 - 1 = 20 \]Simplify it to match the standard form of a circle:\[ (x+1)^2 + (y-1)^2 = 22 \]

Step 5: Identify Circle Properties

The standard form \( (x+1)^2 + (y-1)^2 = 22 \) indicates that the circle has a center at \((-1, 1)\) and a radius of \( \sqrt{22} \).

Step 6: Determine y-coordinates for x = -2 and x = 2

Substitute \( x = -2 \) into \( x^2+2x+y^2-2y = 20 \) to solve for \( y \):\[ (-2)^2 + 2(-2) + y^2 - 2y = 20 \]\[ 4 - 4 + y^2 - 2y = 20 \]\[ y^2 - 2y = 20 \]This simplifies to \( y^2 - 2y - 20 = 0 \). Solve using the quadratic formula to find \( y\).

Step 7: Solve Quadratic for y at x = -2

The equation is \( y^2 - 2y - 20 = 0 \). Using the quadratic formula:\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \( a = 1, b = -2, c = -20 \), leads to:\[ y = \frac{2 \pm \sqrt{4 + 80}}{2} = \frac{2 \pm \sqrt{84}}{2} = \frac{2 \pm 2\sqrt{21}}{2} \]\[ y = 1 \pm \sqrt{21} \]

Step 8: Determine y-coordinates for x = 2

Substitute \( x = 2 \) into the original circle equation:\[ 2^2 + 2\times2 + y^2 - 2y = 20 \]\[ 4 + 4 + y^2 - 2y = 20 \]\[ y^2 - 2y = 12 \]This simplifies to \( y^2 - 2y - 12 = 0 \). Solve this quadratic equation for \( y \).

Step 9: Solve Quadratic for y at x = 2

The equation is \( y^2 - 2y - 12 = 0 \). Using the quadratic formula:\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \( a = 1, b = -2, c = -12 \), leads to:\[ y = \frac{2 \pm \sqrt{4 + 48}}{2} = \frac{2 \pm \sqrt{52}}{2} = \frac{2 \pm 2\sqrt{13}}{2} \]\[ y = 1 \pm \sqrt{13} \]

Step 10: Calculate Distances Between Points

Now, calculate the four possible distances between points with coordinates:1. \((-2, 1+\sqrt{21})\) and \((2, 1+\sqrt{13})\)2. \((-2, 1+\sqrt{21})\) and \((2, 1-\sqrt{13})\)3. \((-2, 1-\sqrt{21})\) and \((2, 1+\sqrt{13})\)4. \((-2, 1-\sqrt{21})\) and \((2, 1-\sqrt{13})\)Each distance can be calculated using the distance formula:\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \].

Key Concepts

Distance Formula

Calculating the distance between two points on a coordinate plane is straightforward with the distance formula. This formula is derived from the Pythagorean Theorem and it calculates the distance

• between the two points,
• given their coordinates

If two points are

• ((x_1, y_1)
• (x_2, y_2)

then the distance between them can be given by: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]This formula calculates the horizontal and vertical distances separately, squares them, adds them up, and then takes the square root of the sum to find the straight-line distance.
In our exercise, we use this fundamental formula to find the lengths between the coordinates

• (-2, 1±\sqrt{21})
• and
• (2, 1±\sqrt{13})

which gives us insight into the various possible distances between the points lying on the circle.

Quadratic Equations

Quadratic equations are polynomial equations of the second degree

• and they play a crucial role in geometry and algebra

The standard form is expressed as: \[ ax^2 + bx + c = 0 \]where

• \(a\), \(b\), and \(c\) are coefficients
• and \(x\) represents the variable

To solve these equations, one commonly uses the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]This can help find the values of \(x\) or \(y\) which satisfy the equation. In our exercise, solving the quadratic equations was fundamental in identifying the possible values for the y-coordinates when the x-values were specified. This allowed us to further explore the intersection of these points with the circle.

Standard Form of a Circle

In geometry, the equation of a circle with a given center and radius can be brought into a particular format known as the "standard form". This form is handy for easily identifying the circle's properties. It is expressed as: \[ (x-h)^2 + (y-k)^2 = r^2 \]here,

• \((h,k)\) is the center of the circle
• and \(r\) is the radius

In our exercise, transforming the given equation into this form was essential. It helped locate the center at

• (-1, 1)
• and determine the radius of \( \sqrt{22} \).

This transformation made it easier to understand and solve the geometric problem of determining distances between specified points on the circle.

Completing the Square

Completing the square is a useful algebraic technique primarily used for solving quadratic equations but also beneficial in transforming equations into the standard form. It helps rearrange equations for easier interpretation.

• The method involves creating a perfect square trinomial from part of the equation

To complete the square for a quadratic expression like \(x^2 + bx \), we rewrite it as: \[ (x + \frac{b}{2})^2 - (\frac{b}{2})^2 \]In the context of the given circle equation, completing the square was essential. It helped convert the equation \[ x^2 + 2x + y^2 - 2y = 20 \]into

• \((x+1)^2 + (y-1)^2 = 22\)

This transformation allows us to easily identify geometric properties such as the circle’s center and radius, facilitating the understanding and solution of the problem.