Problem 32 - Precalculus: Functions and Graphs

Question

Find a polar equation that has the same graph as the equation in \(x\) and \(y\). $$(x+2)^{2}+y^{2}=4$$

Step-by-Step Solution

Answer

The polar equation is \(r = -4 \cos(\theta)\).

Step 1: Understand the Given Equation

We have the Cartesian equation \((x+2)^2 + y^2 = 4\). This represents a circle in the coordinate plane. The circle's center is at \((-2, 0)\) and its radius is \(2\).

Step 2: Express in Polar Coordinates

In polar coordinates, \(x = r \cos(\theta)\) and \(y = r \sin(\theta)\). Substitute these expressions into the given equation: \[((r \cos(\theta) + 2)^2 + (r \sin(\theta))^2 = 4)\].

Step 3: Expand the Equation

Expand \((r \cos(\theta) + 2)^2\) to get \((r \cos(\theta))^2 + 4r \cos(\theta) + 4\), so the equation becomes: \[r^2 \cos^2(\theta) + 4r \cos(\theta) + 4 + r^2 \sin^2(\theta) = 4\]

Step 4: Simplify Using Identities

Use the identity \(r^2 \cos^2(\theta) + r^2 \sin^2(\theta) = r^2\). Simplify to get: \[r^2 + 4r \cos(\theta) + 4 = 4\]

Step 5: Solve for r

Subtract \(4\) from both sides: \[r^2 + 4r \cos(\theta) = 0\]Factor the equation: \[r(r + 4 \cos(\theta)) = 0\]The solutions are \(r = 0\) or \(r = -4 \cos(\theta)\).

Step 6: Choose the Non-trivial Solution

Since \(r = 0\) represents only the origin, which doesn't match the whole circle, we choose the non-trivial solution: \[r = -4 \cos(\theta)\]

Key Concepts

Cartesian Coordinates

In mathematics, Cartesian Coordinates are a way to describe the position of a point in a plane using two numbers. Each point is defined by an

• X-coordinate: which measures the point's distance from the y-axis.
• Y-coordinate: which measures the point's distance from the x-axis.

The notation \((x, y)\) is often used for these coordinates.
In our exercise, we start with a Cartesian equation \((x+2)^2 + y^2 = 4\). This equation is a form of the general circle equation in Cartesian coordinates, where the circle's center is at the point \((-2, 0)\)and its radius is 2.
Moving from Cartesian coordinates to polar coordinates allows us to explore the geometric relationship in a different way.

Circle Equations

Circle equations are used to describe the shape and position of a circle in geometry. The standard form of a circle's equation in Cartesian coordinates is \((x - h)^2 + (y - k)^2 = r^2\).
Here:

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

The exercise starts with the equation \((x+2)^{2}+y^{2}=4\).
This format reveals that the circle has a center at \((-2, 0)\)and a radius \(r=2\). Circles in geometry can also be expressed using polar coordinates, which might be more useful for different types of problems and contexts.

Trigonometric Identities

Trigonometric Identities are equations involving trigonometric functions that hold true for all angles. In the context of the exercise, we used the identity\[r^2 \cos^2(\theta) + r^2 \sin^2(\theta) = r^2\]This identity stems from the Pythagorean identity \(\cos^2(\theta) + \sin^2(\theta) = 1\).
This is useful when simplifying equations because it allows us to convert expressions involving both \(\cos(\theta)\)and \(\sin(\theta)\) to a single term in \(r\).Such identities are critical when converting between Cartesian and polar equations, as they help us reduce and simplify complex expressions efficiently.

Coordinate Conversion

Coordinate Conversion is the process of translating a point's representation from one coordinate system to another. It involves using relationships between coordinates, such as:

• \(x = r \cos(\theta)\)
• \(y = r \sin(\theta)\)

These allow us to express Cartesian coordinates in terms of polar coordinates. The conversion is vital when you need to evaluate geometrical properties more naturally described in a polar system, like circles.
In our exercise, we convert the equation \((x+2)^2 + y^2 = 4\) into polar coordinates. First, substitute \(x = r \cos(\theta)\) and \(y = r \sin(\theta)\) into the Cartesian equation. Simplifying, we find a polar equivalent \[r = -4 \cos(\theta)\]. This polar form gives us the same circle as the original Cartesian equation.