Question

# What is represented by the equation $\left( r\cos \theta -a \right)\left( r-a\cos \theta \right)=0$?

Hint: Put $r\cos \theta =x\ and\ r\sin \theta =y$. Square x and y. You have to prove that x = a, represents a line and prove that the equation also represents the equation of the circle.

Given the equation $\left( r\cos \theta -a \right)\left( r-a\cos \theta \right)=0$…………………. (1)
Now let us consider $r\cos \theta =x\ and\ r\sin \theta =y$.
Now squaring and adding x and y, we get
\begin{align} & {{x}^{2}}+{{y}^{2}}={{\left( r\cos \theta \right)}^{2}}+{{\left( r\sin \theta \right)}^{2}} \\ & \Rightarrow {{x}^{2}}+{{y}^{2}}={{r}^{2}}\left[ {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right] \\ \end{align}
We know,
\begin{align} & {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 \\ & \therefore {{x}^{2}}+{{y}^{2}}={{r}^{2}}...............\left( 2 \right) \\ & \Rightarrow r=\sqrt{{{x}^{2}}+{{y}^{2}}} \\ \end{align}
Now take $\left( r\cos \theta -a \right)\left( r-a\cos \theta \right)=0$
Take, $\left( r\cos \theta -a \right)=0$
$\Rightarrow r\cos \theta =a$
We know $r\cos \theta =x$.
$\therefore x=a$, which represents a vertical straight line.
Take, $\left( r-a\cos \theta \right)=0$.
$\therefore r=a\cos \theta$
Now multiply r on LHS and RHS.
$\Rightarrow {{r}^{2}}=ar\cos \theta$
We know $r\cos \theta =x$.
$\Rightarrow {{r}^{2}}=ax$
Now substitute ${{r}^{2}}$ on equation (2).
\begin{align} & {{x}^{2}}+{{y}^{2}}={{r}^{2}} \\ & {{x}^{2}}+{{y}^{2}}=ax \\ & \Rightarrow {{x}^{2}}+{{y}^{2}}-ax=0 \\ & \left( {{x}^{2}}-ax \right)+{{y}^{2}}=0 \\ \end{align}
Now this represents the equation of a circle.
Add ${{\left( \dfrac{a}{2} \right)}^{2}}$ with $\left( {{x}^{2}}-ax \right)$ and subtract ${{\left( \dfrac{a}{2} \right)}^{2}}$.
\begin{align} & \Rightarrow {{x}^{2}}-ax+{{\left( \dfrac{a}{2} \right)}^{2}}+{{y}^{2}}{{\left( \dfrac{a}{2} \right)}^{2}}=0 \\ & \Rightarrow {{\left( x-\dfrac{a}{2} \right)}^{2}}+{{y}^{2}}={{\left( \dfrac{a}{2} \right)}^{2}} \\ \end{align}
This is of the form of a circle with centre $\left( \dfrac{a}{2},0 \right)$ and radius $\dfrac{a}{2}$.
Thus, the combined equation represents a circle and a straight line.
$\therefore x=a$ represents a straight line
and ${{\left( x-\dfrac{a}{2} \right)}^{2}}+{{y}^{2}}={{\left( \dfrac{a}{2} \right)}^{2}}$represents equation of a circle.

Note: Take $r\cos \theta =x\ and\ r\sin \theta =y$ to get the required results. In the question ‘r’ represents the radius of the circle. If you consider $x=a\cos \theta$ won’t provide desirable results.