Question

Show that the graph of the equation $r=-2a\cos \theta$, $a>0$

is a circle of radius a with center at $(-a,0)$in rectangular

coordinates.

Short answer

${(x+a)}^2+y^2=a^2$

Step-by-step solution

Step 1. Given

given polar coordinate

$r=-2a\cos \theta$

Step 2. Concept

To convert given polar coordinate into rectangular coordinate we try to deduce the given polar coordinate into the following expression:

$x^2+y^2=2ax$

Step 3. Calculation

Consider the given polar coordinate

$$\begin{gathered} r=-2a\cos \theta \\ multiplytheaboveequationbyr \\ {r}^2=-2a.r\cos \theta \\ substitute{x}^2+y^{2}inplaceof{r}^{2}andr\cos \theta forxweget \\ {x}^2+y^2=-2ax \\ {x}^2+y^2+2ax=0 \\ Nowadd{a}^{2}bothsidestomakeitaperfectsquare \\ {y}^2+(x^2+2ax+a^2)=a^2 \\ {y}^2+{(x+a)}^2=a^{2}whichisoftheformofequationofacircle. \end{gathered}$$