Problem 88 - Algebra and Trigonometry

Question

Show that the graph of the equation \(r=-2 a \cos \theta, a>0\) is a circle of radius \(a\) with center \((-a, 0)\) in rectangular coordinates.

Step-by-Step Solution

Answer

The graph is a circle with center at (-a, 0) and radius a.

Step 1 - Convert Polar Equation to Rectangular Form

Given the polar equation: \[r = -2a \cos \theta\]Use the polar to rectangular coordinate conversions: \[x = r \cos \theta\] and \[y = r \sin \theta\].

Step 2 - Express rcos(θ) in terms of x

Multiply both sides of the polar equation by \(\cos \theta\): \[r \cos \theta = -2a \cos^2 \theta\]Using the identity \(x = r \cos \theta\), we get: \[x = -2a \left(\frac{x}{r}\right)^2\]

Step 3 - Substitute for r

Recall that \(r^2 = x^2 + y^2\). Substitute this into the equation: \[x = -2a \frac{x^2}{x^2 + y^2}\].

Step 4 - Simplify the equation

Multiply both sides by \(x^2 + y^2\): \[x \left( x^2 + y^2 \right) = -2ax^2\].Simplify the equation: \[x^3 + xy^2 = -2a x^2\].

Step 5 - Factor the equation

Rearrange terms to isolate terms involving \(x\) and factor out \(x\):\[x^3 + xy^2 + 2ax^2 = 0\].Factor out \(x\): \[x(x^2 + y^2 + 2ax) = 0\].

Step 6 - Solve for the Circle Equation

The factor \(x=0\) is not relevant for our circle. We focus on the remaining part: \[x^2 + y^2 + 2ax = 0\].Complete the square to reveal the circle's standard form: \[(x+a)^2 + y^2 = a^2\].

Step 7 - Identify Circle's Parameters

The equation \[(x+a)^2 + y^2 = a^2\]describes a circle with center \((-a, 0)\) and radius \(a\).

Key Concepts

polar coordinates

Polar coordinates are a way to describe a point in the plane by its distance from a reference point and the angle from a reference direction. This system contrasts with rectangular (or Cartesian) coordinates, where points are defined by \(x\) and \(y\) values directly. In polar notation:
\(r\) represents the radius, or the distance from the origin.
\( \theta \) represents the angle in degrees or radians, counterclockwise from the positive \(x\)-axis.
The conversion between polar and rectangular coordinates involves trigonometric relationships:
\[ x = r \cos \( \theta \), \quad y = r \sin \( \theta \). \] These relationships allow us to switch from polar coordinates, which describe distance and direction, to rectangular coordinates, which describe location in a grid.

rectangular coordinates

Rectangular coordinates, often known as Cartesian coordinates, identify points in a plane using two perpendicular axes: the x-axis (horizontal) and the y-axis (vertical). Every point is expressed as \( (x, y) \) where:
\(x\) represents the horizontal distance from the origin.
\(y\) represents the vertical distance from the origin.
These coordinates allow for straightforward addition, subtraction, and plotting of points on a grid. The transformation from polar to rectangular coordinates converts angular and radial measurement into this structured grid format, facilitating easier algebraic manipulation and visualization of geometric shapes, like circles.

circle equation

An equation representing a circle in rectangular coordinates generally takes the form:
\[ (x-h)^2 + (y-k)^2 = r^2 \] Here:

• \(h\) and \(k\)
  are the coordinates of the circle's center.
• \(r\)
  is the radius of the circle.

In the given problem, after converting the polar equation \( r = -2a \cos \( \theta \)\) to rectangular form and simplifying, we recognized a standard circle equation. This transformation is crucial because it reveals the geometric figure described by the polar equation, making it easier to visualize and analyze.

complete the square

Completing the square is a method used to rewrite a quadratic equation in a 'perfect square' form, which is beneficial for identifying conic sections, like circles. The general process is:

• Start with a quadratic equation, typically in the form: \[ ax^2 + bx + c = 0 \]
• Isolate the \(x\) terms.
• Add and subtract the necessary constant to form a perfect square trinomial.
• Rewrite the trinomial as a squared binomial.

In the context of our exercise, we transformed
\[ x^2 + y^2 + 2ax = 0 \] into
\[ (x+a)^2 + y^2 = a^2 \] by completing the square. This reveals the circle's center and radius directly, showing the circle's standard form in rectangular coordinates.