Answer
(a) $x^2+y^2=4$ matches (iv) $r=2$
(b) $x^2+(y-1)^2=1$ matches (iii) $r=2\sin \theta$
(c) $x^2-y^2=4$ matches (i) ${r^2}\left( {1 - 2{{\sin }^2}\theta } \right) = 4$
(d) $x+y=4$ matches (ii) $r\left( {\cos \theta + \sin \theta } \right) = 4$
Work Step by Step
(a) We have $x = r\cos \theta $ and $y = r\sin \theta $. So,
$x^2+y^2=4$,
$\begin{array}{l}
{r^2}{\cos ^2}\theta + {r^2}{\sin ^2}\theta = 4\\
{r^2} = 4
\end{array}$
The answer is (iv) $r=2$.
(b) We have $x = r\cos \theta $ and $y = r\sin \theta $. So,
$x^2+(y-1)^2=1$,
$\begin{array}{l}
{r^2}{\cos ^2}\theta + {\left( {r\sin \theta - 1} \right)^2} = 1\\
{r^2}{\cos ^2}\theta + {r^2}{\sin ^2}\theta - 2r\sin \theta + 1 = 1\\
{r^2} - 2r\sin \theta = 0\\
r\left( {r - 2\sin \theta } \right) = 0
\end{array}$
The answer is (iii) $r=2\sin \theta$.
(c) We have $x = r\cos \theta $ and $y = r\sin \theta $. So,
$x^2-y^2=4$,
$\begin{array}{l}
{r^2}{\cos ^2}\theta - {r^2}{\sin ^2}\theta = 4\\
{r^2}\left( {{{\cos }^2}\theta - {{\sin }^2}\theta } \right) = 4
\end{array}$
Since ${\cos ^2}\theta - {\sin ^2}\theta = 1 - 2{\sin ^2}\theta $, the answer is (i) ${r^2}\left( {1 - 2{{\sin }^2}\theta } \right) = 4$.
(d) We have $x = r\cos \theta $ and $y = r\sin \theta $. So,
$x+y=4$,
$r \cos \theta + r \sin \theta=4$
The answer is (ii) $r\left( {\cos \theta + \sin \theta } \right) = 4$.
