Question Number 28399 by ajfour last updated on 25/Jan/18
Answered by ajfour last updated on 25/Jan/18
![eq. of circle: x=r+rcos θ , y=r+rsin θ eq. of line: (x/a)+(y/b)=1 intersection points: θ_1 , θ_2 obeys ((1+cos θ)/a)+((1+sin θ)/b)=(1/r) ((cos θ)/a)+((sin θ)/b)=(1/r)−((a+b)/(ab)) let △θ=θ_2 −θ_1 ⇒ bcos ((△θ)/2)cos (((θ_1 +θ_2 )/2))+ asin (((θ_1 +θ_2 )/2))cos ((△θ)/2)=((ab)/r)−(a+b) Since tan (((θ_1 +θ_2 )/2))=(a/b) (b^2 /( (√(a^2 +b^2 ))))cos ((△θ)/2)+(a^2 /( (√(a^2 +b^2 )))) cos ((△θ)/2)=((ab)/r)−(a+b) (√(a^2 +b^2 ))cos (((△θ)/2))=((ab)/r)−(a+b) ((△θ)/2)=cos^(−1) [(1/( (√(a^2 +b^2 ))))(((ab)/r)−a−b)] Required Area= ((r^2 △θ)/2)−(1/2)×(2rsin ((△θ)/2))(rcos ((△θ)/2)) or A=((r^2 △θ)/2)(1−((sin △θ)/(△θ))) .](https://www.tinkutara.com/question/Q28400.png)
$${eq}.\:{of}\:{circle}: \\ $$$${x}={r}+{r}\mathrm{cos}\:\theta\:\:,\:{y}={r}+{r}\mathrm{sin}\:\theta \\ $$$${eq}.\:{of}\:{line}: \\ $$$$\frac{{x}}{{a}}+\frac{{y}}{{b}}=\mathrm{1} \\ $$$$\:{intersection}\:{points}:\:\:\theta_{\mathrm{1}} ,\:\theta_{\mathrm{2}} \:{obeys} \\ $$$$\frac{\mathrm{1}+\mathrm{cos}\:\theta}{{a}}+\frac{\mathrm{1}+\mathrm{sin}\:\theta}{{b}}=\frac{\mathrm{1}}{{r}} \\ $$$$\frac{\mathrm{cos}\:\theta}{{a}}+\frac{\mathrm{sin}\:\theta}{{b}}=\frac{\mathrm{1}}{{r}}−\frac{{a}+{b}}{{ab}} \\ $$$${let}\:\:\bigtriangleup\theta=\theta_{\mathrm{2}} −\theta_{\mathrm{1}} \\ $$$$\Rightarrow\:{b}\mathrm{cos}\:\frac{\bigtriangleup\theta}{\mathrm{2}}\mathrm{cos}\:\left(\frac{\theta_{\mathrm{1}} +\theta_{\mathrm{2}} }{\mathrm{2}}\right)+ \\ $$$$\:\:{a}\mathrm{sin}\:\left(\frac{\theta_{\mathrm{1}} +\theta_{\mathrm{2}} }{\mathrm{2}}\right)\mathrm{cos}\:\frac{\bigtriangleup\theta}{\mathrm{2}}=\frac{{ab}}{{r}}−\left({a}+{b}\right) \\ $$$${Since}\:\:\mathrm{tan}\:\left(\frac{\theta_{\mathrm{1}} +\theta_{\mathrm{2}} }{\mathrm{2}}\right)=\frac{{a}}{{b}} \\ $$$$\frac{{b}^{\mathrm{2}} }{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\mathrm{cos}\:\frac{\bigtriangleup\theta}{\mathrm{2}}+\frac{{a}^{\mathrm{2}} }{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\:\mathrm{cos}\:\:\frac{\bigtriangleup\theta}{\mathrm{2}}=\frac{{ab}}{{r}}−\left({a}+{b}\right) \\ $$$$\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\mathrm{cos}\:\left(\frac{\bigtriangleup\theta}{\mathrm{2}}\right)=\frac{{ab}}{{r}}−\left({a}+{b}\right) \\ $$$$\frac{\bigtriangleup\theta}{\mathrm{2}}=\mathrm{cos}^{−\mathrm{1}} \left[\frac{\mathrm{1}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\left(\frac{{ab}}{{r}}−{a}−{b}\right)\right] \\ $$$${Required}\:{Area}= \\ $$$$\:\:\:\frac{{r}^{\mathrm{2}} \bigtriangleup\theta}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}×\left(\mathrm{2}{r}\mathrm{sin}\:\frac{\bigtriangleup\theta}{\mathrm{2}}\right)\left({r}\mathrm{cos}\:\frac{\bigtriangleup\theta}{\mathrm{2}}\right) \\ $$$${or}\:\:{A}=\frac{{r}^{\mathrm{2}} \bigtriangleup\theta}{\mathrm{2}}\left(\mathrm{1}−\frac{\mathrm{sin}\:\bigtriangleup\theta}{\bigtriangleup\theta}\right)\:. \\ $$
Commented by mrW2 last updated on 25/Jan/18
![I would try like this: ((cos θ)/a)+((sin θ)/b)=(1/r)−((a+b)/(ab)) ⇒a sin θ+b cos θ=((ab)/r)−a−b (a/( (√(a^2 +b^2 )))) sin θ+(b/( (√(a^2 +b^2 )))) cos θ=(1/( (√(a^2 +b^2 ))))(((ab)/r)−a−b) sin α sin θ+cos α cos θ=(1/( (√(a^2 +b^2 ))))(((ab)/r)−a−b) with α=tan^(−1) (a/b) ⇒cos (θ−α)=(1/( (√(a^2 +b^2 ))))(((ab)/r)−a−b) θ=tan^(−1) (a/b)±cos^(−1) [(1/( (√(a^2 +b^2 ))))(((ab)/r)−a−b)] Δθ=θ_2 −θ_1 =2 cos^(−1) [(1/( (√(a^2 +b^2 ))))(((ab)/r)−a−b)] ...... A=(r^2 /2)(Δθ−sin Δθ)](https://www.tinkutara.com/question/Q28409.png)
$${I}\:{would}\:{try}\:{like}\:{this}: \\ $$$$\frac{\mathrm{cos}\:\theta}{{a}}+\frac{\mathrm{sin}\:\theta}{{b}}=\frac{\mathrm{1}}{{r}}−\frac{{a}+{b}}{{ab}} \\ $$$$\Rightarrow{a}\:\mathrm{sin}\:\theta+{b}\:\mathrm{cos}\:\theta=\frac{{ab}}{{r}}−{a}−{b} \\ $$$$\frac{{a}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\:\mathrm{sin}\:\theta+\frac{{b}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\:\mathrm{cos}\:\theta=\frac{\mathrm{1}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\left(\frac{{ab}}{{r}}−{a}−{b}\right) \\ $$$$\mathrm{sin}\:\alpha\:\mathrm{sin}\:\theta+\mathrm{cos}\:\alpha\:\mathrm{cos}\:\theta=\frac{\mathrm{1}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\left(\frac{{ab}}{{r}}−{a}−{b}\right) \\ $$$${with}\:\alpha=\mathrm{tan}^{−\mathrm{1}} \frac{{a}}{{b}} \\ $$$$\Rightarrow\mathrm{cos}\:\left(\theta−\alpha\right)=\frac{\mathrm{1}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\left(\frac{{ab}}{{r}}−{a}−{b}\right) \\ $$$$\theta=\mathrm{tan}^{−\mathrm{1}} \frac{{a}}{{b}}\pm\mathrm{cos}^{−\mathrm{1}} \left[\frac{\mathrm{1}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\left(\frac{{ab}}{{r}}−{a}−{b}\right)\right] \\ $$$$\Delta\theta=\theta_{\mathrm{2}} −\theta_{\mathrm{1}} =\mathrm{2}\:\mathrm{cos}^{−\mathrm{1}} \left[\frac{\mathrm{1}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\left(\frac{{ab}}{{r}}−{a}−{b}\right)\right] \\ $$$$…… \\ $$$${A}=\frac{{r}^{\mathrm{2}} }{\mathrm{2}}\left(\Delta\theta−\mathrm{sin}\:\Delta\theta\right)\: \\ $$
Commented by ajfour last updated on 25/Jan/18
$${thanks}\:{very}\:{much}\:{Sir},\:{i}\:{corrected}\:. \\ $$