Problem 25 - Thomas Calculus

Question

Find polar equations for the circles in Exercises \(21-28 .\) Sketch each circle in the coordinate plane and label it with both its Cartesian and polar equations. $$ x^{2}+2 x+y^{2}=0 $$

Step-by-Step Solution

Answer

The polar equation for the circle is \(r = -2 \cos \theta\).

Step 1: Complete the square for the x-terms

The given Cartesian equation of the circle is \(x^{2} + 2x + y^{2} = 0\). To put this into a more recognizable form of the circle equation \((x-h)^2 + (y-k)^2 = r^2\), we first need to complete the square for the \(x\) terms:\[x^{2} + 2x = (x + 1)^{2} - 1\]So, the equation becomes \((x + 1)^{2} - 1 + y^{2} = 0\).

Step 2: Rearrange to get standard form of a circle

Now, move constant terms to the other side of the equation:\[(x + 1)^{2} + y^{2} = 1\]This equation is in the standard form of a circle with center \((-1, 0)\) and radius \(1\).

Step 3: Convert to polar coordinates

To convert the Cartesian equation \((x + 1)^2 + y^2 = 1\) into polar coordinates, use the substitutions \(x = r \cos \theta\) and \(y = r \sin \theta\). Substitute these into the equation:\[(r \cos \theta + 1)^{2} + (r \sin \theta)^{2} = 1\]Expanding \((r \cos \theta + 1)^{2}\):\[r^{2} \cos^{2} \theta + 2r \cos \theta + 1 + r^{2} \sin^{2} \theta = 1\]Combine terms with \(r^2\):\[r^{2} (\cos^{2} \theta + \sin^{2} \theta) + 2r \cos \theta + 1 = 1\]Since \(\cos^{2} \theta + \sin^{2} \theta = 1\), simplify to:\[r^{2} + 2r \cos \theta + 1 = 1\]Solve for \(r\):\[r^{2} + 2r \cos \theta = 0\]Factor out \(r\):\[r (r + 2 \cos \theta) = 0\]Thus, \(r = 0\) or \(r = -2 \cos \theta\).

Step 4: Sketch the circle and label equations

The circle has center \((-1, 0)\) and radius \(1\), so draw a circle centered at \((-1, 0)\) with radius \(1\) on the Cartesian coordinate plane. Label the Cartesian equation as \((x + 1)^2 + y^2 = 1\) and the polar equation as \(r = -2 \cos \theta\).

Key Concepts

Cartesian Coordinates

Understanding Cartesian coordinates is fundamental in graphing equations such as circles. The Cartesian coordinate system is based on two intersecting lines: the x-axis (horizontal) and the y-axis (vertical).

• Each point in this system is defined by a pair of numbers (x, y).
• The x-value indicates the position along the horizontal axis, while the y-value indicates the position along the vertical axis.

For the given problem, the Cartesian equation of the circle is given as \( x^2 + 2x + y^2 = 0 \). This can be reformed into the standard circle equation format through a method called 'completing the square'.Such equations generally have patterns known as \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) is the circle's center, and \(r\) is the radius.

Circle Equations

A circle equation represents all points that are equidistant from a center point. The classic form is \((x-h)^2 + (y-k)^2 = r^2\), describing a circle with center \((h, k)\) and radius \(r\).Using completing the square, the original equation \(x^2 + 2x + y^2 = 0\) is rearranged into: \((x + 1)^2 + y^2 = 1\).

• Here, the circle's center is located at \((-1, 0)\).
• The radius is \(1\), since our final equation shows \((x + 1)^2 + y^2 = 1\).

This standard form makes it easy to sketch the circle's position and size on the coordinate plane, providing a clear visual representation of the circle's geometry.

Coordinate Conversion

Converting from Cartesian to polar coordinates involves using trigonometric relationships. In the polar coordinate system, a point's position is described by a radius \(r\) and an angle \(\theta\) from the positive x-axis.

• To convert, use: \(x = r \cos \theta\) and \(y = r \sin \theta\).

For our converted circle equation \((x + 1)^2 + y^2 = 1\), substitute these expressions:\[(r \cos \theta + 1)^2 + (r \sin \theta)^2 = 1\]Simplifying using the identity \(\cos^2 \theta + \sin^2 \theta = 1\), gives us:\[r^2 + 2r \cos \theta = 0\]This equation factors to \(r(r + 2 \cos \theta) = 0\), revealing two possible solutions: \(r = 0\), or \(r = -2 \cos \theta\), the latter describing the circle.

Completing the Square

Completing the square is a technique to transform quadratic equations into a standard form by creating a perfect square trinomial.Let's see how it's applied to this problem:The original equation is \( x^2 + 2x + y^2 = 0 \). To complete the square for the \(x\) terms \(x^2 + 2x\):

• Take half of the \(x\) term coefficient (which is 2), yielding 1.
• Square it to obtain 1.
• Add and subtract this square inside the equation: \((x + 1)^2 - 1 + y^2 = 0\).

This results in the rearranged circle equation: \((x + 1)^2 + y^2 = 1\), with center \((-1, 0)\) and radius 1. Completing the square turns complex expressions into simpler forms, making both understanding and solving them easier.