Question

$\Huge \fbox{{\color{magenta}{\textsf {\textbf {Question :-}}}}}$

Find the length of the arc x² + y² + 2ax = 0 in the second quadrant​

Answer 1

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

Given equation of circle is

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

Let reduced this general equation of circle to standard form.

On rearranging can be rewritten as

$\rm :\longmapsto\: ({x}^{2} + 2ax) + {y}^{2} = 0$

$\rm :\longmapsto\: ({x}^{2} + 2ax + {a}^{2} - {a}^{2} ) + {y}^{2} = 0$

$\rm :\longmapsto\: {(x + a)}^{2} + {y}^{2} - {a}^{2} = 0$

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

It implies, Center of circle = ( - a, 0 ) and radius is a units.

Now,

We have,

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

Differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx} {(x + a)}^{2} + \dfrac{d}{dx}{y}^{2} = \dfrac{d}{dx} {a}^{2}$

$\rm :\longmapsto\:2(x + a) + 2y\dfrac{dy}{dx} = 0$

$\rm :\longmapsto\:(x + a) + y\dfrac{dy}{dx} = 0$

$\bf\implies \:\dfrac{dy}{dx} \: = \: - \: \dfrac{x + a}{y}$

Now,

For the length of arc of circle, in second quadrant, the values of x varies from x = - 2a to x = 0

Now, we know that

$\rm :\longmapsto\:Length \: of \: arc = \displaystyle\int^{0}_{ - 2a} \sqrt{1 + {\bigg(\dfrac{dy}{dx} \bigg) }^{2} }dx$

So, on substituting the value, we get

$\rm \:  \:  =  \:  \: \displaystyle\int^{0}_{ - 2a} \sqrt{1 + {\bigg(\dfrac{ - (a + x)}{y} \bigg) }^{2} }dx$

$\rm \:  \:  =  \:  \: \displaystyle\int^{0}_{ - 2a} \sqrt{1 + {\bigg(\dfrac{a + x}{y} \bigg) }^{2} }dx$

$\rm \:  \:  =  \:  \: \displaystyle\int^{0}_{ - 2a} \sqrt{1 + \dfrac{ {(a + x)}^{2} }{ {y}^{2} } }dx$

$\rm \:  \:  =  \:  \: \displaystyle\int^{0}_{ - 2a} \sqrt{ \dfrac{ {y}^{2} + {(a + x)}^{2} }{ {y}^{2} } }dx$

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

$\rm \:  \:  =  \:  \: \displaystyle\int^{0}_{ - 2a} \sqrt{ \dfrac{ {a}^{2} }{ {y}^{2} } }dx$

$\rm \:  \:  =  \:  \: \displaystyle\int^{0}_{ - 2a} \sqrt{ \dfrac{ {a}^{2} }{ {a}^{2} - {(x + a)}^{2} } }dx$

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

$\rm \:  \:  =  \:  \: \displaystyle\int^{0}_{ - 2a} \frac{a}{ \sqrt{ {a}^{2} - {(x + a)}^{2} } } dx$

$\rm \: = \:a\bigg[ {sin}^{ - 1}\dfrac{a + x}{a} \bigg]^{0}_{ - 2a}$

$\rm \: = \:a\bigg[ {sin}^{ - 1}\dfrac{a + 0}{a} - {sin}^{ - 1} \dfrac{a - 2a}{a} \bigg]$

$\rm \: = \:a\bigg[ {sin}^{ - 1}\dfrac{a}{a} - {sin}^{ - 1} \dfrac{- a}{a} \bigg]$

$\rm \: = \:a\bigg[ {sin}^{ - 1}(1) + {sin}^{ - 1} (1) \bigg]$

$\rm \: = \:a\bigg[ 2{sin}^{ - 1}(1) \bigg]$

$\rm \: = \:a \times 2 \times \dfrac{\pi}{2}$

$\rm \: = \:a \pi \: units.$