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

Answer

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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.

Complete step-by-step answer:

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.

Complete step-by-step answer:

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.

Last updated date: 30th Sep 2023

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