Question

Convert the rectangular equation to polar form.
${x}^{2} + (y - 2) {}^{2} = 4$
​

Answer 1

Given :-

$\sf {x}^{2} + (y - 2) {}^{2} = 4$

To Find :-

The Equation in polar form

Solution :-

We knows a algebraic Identity i.e

$\quad \qquad { \bigstar { \underline { \boxed { \bf { \red { ( a - b)² = a² + b² - 2ab }}}}}}{\bigstar}$

Using this the given equation transforms to ;

$\quad { : \longmapsto { \sf { x² + { \bigg [ ( y² ) + ( 2² ) - 2 × 2 × y }\bigg ] = 4 }}}$

$\quad { : \longmapsto { \sf { x² + y² - 4y = 4 - 4 }}}$

$\quad { : \longmapsto { \sf{ x² + y² - 4y = 0 }}}$

Now , To Change the equation to polar form Put ;

• $\sf r Cos \theta = x$

• $\sf r Sin \theta = y$

For all $\sf r \in \mathbb N$

Now , The given Equation transforms to ;

$\quad { : \longmapsto { \sf { ( r Cos \theta )² + ( r Sin \theta )² = 4y }}}$

$\quad { : \longmapsto { \sf { r² Cos² \theta + r² Sin² \theta = 4y }}}$

$\quad { : \longmapsto { \sf { r² ( Sin² \theta + Cos² \theta ) = 4y }}}$

Now , we knows a Identity i.e ;

$\quad \qquad { \bigstar { \underline { \boxed { \bf { \red { ( Sin² \theta + Cos² \theta ) = 1 }}}}}}{\bigstar}$

Using this we have ;

$\quad { : \longmapsto { \sf { r² ( 1 ) = 4y }}}$

$\quad { : \longmapsto { \sf { r² = 4y }}}$

$\quad { : \longmapsto { \sf { r² = ( \pm \sqrt{4y} ) }}}$

$\quad { : \longmapsto { \sf { ( r )² = ( \pm 2\sqrt{y} )²}}}$

Square gets cancelled & then we have ;

$\quad { : \longmapsto { \sf { r = \pm 2 \sqrt{y} }}}$

$\quad { \sf { \sf But \:\: \sf r \neq - 2\sqrt{y} }} \quad \qquad { \bigg [ \because r \in \mathbb N } \bigg ]$

$\quad \qquad { \bigstar { \underline { \boxed { \bf { \red { \therefore r = 2\sqrt{y} }}}}}}{\bigstar}$

For the graph of the equation kindly see the attachment !

Answer 2

$\large\underline{\sf{Solution-}}$

Given equation of circle is

$\rm :\longmapsto\: {x}^{2} + {(y - 2)}^{2} = 4$

can be rewritten as

$\rm :\longmapsto\: {x}^{2} + {y}^{2} + 4 - 4y = 4$

$\: \: \: \: \: \: \: \: \red{\bigg \{ \because \: {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy \bigg \}}$

$\rm :\longmapsto\: {x}^{2} + {y}^{2} - 4y = 0$

We know, Polar coordinates is represented as

$\red{\rm :\longmapsto\:(r, \: \theta )\:}$

where,

$\rm :\longmapsto\:\boxed{ \tt{ \: r = \sqrt{ {x}^{2} + {y}^{2}} \: }}$

and

$\rm :\longmapsto\:\boxed{ \tt{ \: \theta = {tan}^{ - 1} \frac{y}{x} \: }}$

And, Further more

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: x = r \: cos\theta \: }}}$

and

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: y = r \: sin\theta \: }}}$

So, on plug in all these values, we get

$\rm :\longmapsto\: {r}^{2} - 4rsin\theta = 0$

$\rm :\longmapsto\:r(r - 4sin\theta) = 0$

$\bf\implies \:r = 0 \: \: or \: \: r = 4 \: sin\theta$

is the required equation of circle in polar coordinates.