Question

Which of the following equations is of circle?

• A. $r=2\sin \theta$
• B. $r^2\cos 2\theta =1$
• C. $r(4\cos \theta +5\sin \theta )=3$
• D. $5=r(1+\sqrt{2}\cos \theta )$

Answer 1

The correct option is A

$r=2\sin \theta$

Explanation for correct option:

For option (a)

Given equation $r=2\sin \theta$………(1)

We have: $x=r\cos \theta$ and $y=r\sin \theta$
Multiplying both sides of equation (1) by $r$ and we get

$r^2=2r\sin \theta$

$\Rightarrow r^2=2y$ $\left[\because y=r\sin \theta \right]$
Multiplying both sides of equation (1) by $\left({\sin }^2\theta +{\cos }^2\theta \right)$ and we get

$\Rightarrow r^2\left({\sin }^2\theta +{\cos }^2\theta \right)=2y\left({\sin }^2\theta +{\cos }^2\theta \right)$

$\Rightarrow r^2{\sin }^2\theta +r^2{\cos }^2\theta =2y\times 1$ $\left[\because {\sin }^2\theta +{\cos }^2\theta =1\right]$

$\Rightarrow {\left(r\sin \theta \right)}^2+{\left(r\cos \theta \right)}^2=2y\times 1$

$$\begin{gathered} \Rightarrow {\left(y\right)}^2+{\left(x\right)}^2=2y\times 1 \\ \Rightarrow x^2+y^2=2y \end{gathered}$$

This is an equation of a circle.

Explanation for incorrect option:

For option (b)

Given $r^2\cos 2\theta =1$

Now

$r^2\cos 2\theta =1$

$\Rightarrow r^2\left({\cos }^2\theta -{\sin }^2\theta \right)=1$ $\left[\because {\cos }^2\theta -{\sin }^2\theta =\cos 2\theta \right]$

$\Rightarrow r^2{\cos }^2\theta -r^2{\sin }^2\theta =1$

$\Rightarrow x^2-y^2=1$

This is not an equation of a circle.

For option (c)

Given $r(4\cos \theta +5\sin \theta )=3$
Now

$r(4\cos \theta +5\sin \theta )=3$

$$\begin{gathered} \Rightarrow 4r\cos \theta +5r\sin \theta =3 \\ \Rightarrow 4x+5y=3 \end{gathered}$$

This is a linear equation.

Therefore this is not an equation of a circle.

For option (d)

Given $5=r(1+\sqrt{2}\cos \theta )$

Now

$5=r(1+\sqrt{2}\cos \theta )$

$\Rightarrow 5=r+\sqrt{2}r\cos \theta$

$\Rightarrow 5=\sqrt{x^2+y^2}+\sqrt{2}x$ $\left[\because x^2+y^2=r^2\right]$

This is not an equation of a circle.
Hence, option (a) is the correct option.