Problem 78 - Algebra and Trigonometry

Question

Show that each statement is true by converting the given polar equation to a rectangular equation. Show that the graph of \(r=a \cos \theta\) is a circle with center at \(\left(\frac{a}{2}, 0\right)\) and radius \(\frac{a}{2}\)

Step-by-Step Solution

Answer

The graph of the polar equation \(r=a \cos \theta\) corresponds to a circle in rectangular coordinates with center \((a/2, 0)\) and radius \(a/2\).

Step 1: Conversion from Polar to Rectangular Coordinates

Start with the given polar equation \(r=a \cos \theta\). Multiply both sides by \(r\) to get \(r^2 = ar \cos \theta\). Now, convert \(r^2\) and \(r \cos \theta\) to rectangular form using the relationships \(r^2 = x^2 + y^2\) and \(r \cos \theta = x\). This yields the equation \(x^2 + y^2 = ax\).

Step 2: Reformatting the Equation

Rearranging the above equation by moving \(ax\) to the left-hand side, we get \(x^2 - ax + y^2 = 0\). The goal is to complete the square for the \(x\) terms, which means adding \((a/2)^2\) to both sides of the equation to balance it out. The equation becomes \(x^2 - ax + (a/2)^2 + y^2 = (a/2)^2\). Simplify this to get \((x - a/2)^2 + y^2 = (a/2)^2\), which is the standard form of a circle's equation.

Step 3: Verification of Circle's Center and Radius

Comparing the equation \((x - a/2)^2 + y^2 = (a/2)^2\) with the standard form \((x-h)^2 + (y-k)^2 = r^2\), it can be seen that the circle is centered at \((h, k) = (a/2, 0)\) and has a radius of \(r = a/2\), thus validating the given statement.