Question

for a positive real number a, the graph of which of the following polar equations is a circle with radius a and center at (a,0) in rectangular coordinates?
choose the correct answer below
r = -2a cos θ
r = -2a sin θ
r = 2a sin θ
r = 2a cos θ

Explanation:

Step1: Recall the conversion formula

The conversion from polar coordinates \((r,\theta)\) to rectangular coordinates \((x,y)\) is \(x = r\cos\theta\), \(y=r\sin\theta\), and \(r^{2}=x^{2}+y^{2}\). The general form of a circle in rectangular coordinates is \((x - h)^{2}+(y - k)^{2}=R^{2}\), where \((h,k)\) is the center and \(R\) is the radius. For a circle with center \((a,0)\) and radius \(a\), its rectangular - coordinate equation is \((x - a)^{2}+y^{2}=a^{2}\). Expand it: \(x^{2}-2ax + a^{2}+y^{2}=a^{2}\), so \(x^{2}+y^{2}=2ax\).

Step2: Substitute \(x = r\cos\theta\) and \(r^{2}=x^{2}+y^{2}\)

Substitute \(x = r\cos\theta\) and \(r^{2}=x^{2}+y^{2}\) into \(x^{2}+y^{2}=2ax\). We get \(r^{2}=2ar\cos\theta\). Since \(r
eq0\) (if \(r = 0\) is a special case that can be considered separately, but for the non - zero part of the polar curve), divide both sides of the equation \(r^{2}=2ar\cos\theta\) by \(r\).

Answer:

\(r = 2a\cos\theta\)