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Question Number 3877 by Rasheed Soomro last updated on 23/Dec/15

What is the area of overlapping region of two circles having radii r_1 and r_2 when the distance between their centres is c, given that r_1 +r_2 >c.

$$\mathcal{W}{hat}\:{is}\:{the}\:{area}\:{of}\:\:{overlapping} \\ $$ $${region}\:{of}\:{two}\:{circles}\:{having}\:{radii} \\ $$ $$\boldsymbol{\mathrm{r}}_{\mathrm{1}} \:{and}\:\boldsymbol{\mathrm{r}}_{\mathrm{2}} \:{when}\:{the}\:{distance}\:{between} \\ $$ $${their}\:{centres}\:{is}\:\:\boldsymbol{\mathrm{c}},\:{given}\:{that}\:\boldsymbol{\mathrm{r}}_{\mathrm{1}} +\boldsymbol{\mathrm{r}}_{\mathrm{2}} >\boldsymbol{\mathrm{c}}. \\ $$

Commented byprakash jain last updated on 24/Dec/15

I will make the correction. The above procedure is for centers on different side of common chord.

$$\mathrm{I}\:\mathrm{will}\:\mathrm{make}\:\mathrm{the}\:\mathrm{correction}. \\ $$ $$\mathrm{The}\:\mathrm{above}\:\mathrm{procedure}\:\mathrm{is}\:\mathrm{for}\:\mathrm{centers}\:\mathrm{on}\:\mathrm{different} \\ $$ $$\mathrm{side}\:\mathrm{of}\:\mathrm{common}\:\mathrm{chord}. \\ $$

Commented byprakash jain last updated on 24/Dec/15

Common chord AB center O_1 and O_2 O_1 O_2 cuts AB at X ∠AO_1 X=α ∠AO_2 X=β AB=2r_1 sin α=2r_2 sin β O_1 X=r_1 cos α, XO_2 =r_2 cos β r_1 cos α+r_2 cos β=c area of △AO_1 B=(1/2)AB×O_1 X=r_1 ^2 cos αsin α area of △AO_2 B=(1/2)AB×O_2 X=r_2 ^2 cos βsin β area of overlapin portion=area of circle segment − area of triangle Area of overlapping portion on side of circle with radius r_1 =αr_1 −area △AO_1 B X=(1/2)αr_1 ^2 −(1/2)r_1 ^2 sin 2α ...(1) Area of overlapping portion on side of circle with radius r_2 =βr_2 −area △AO_2 B Y=(1/2)βr_2 ^2 −(1/2)r_2 ^2 sin 2β ...(2) Next step is to find α, β r_1 cos α+r_2 cos β=c⇒r_1 ^2 cos^2 α=c^2 −2cr_2 cos β+r_2 ^2 cos^2 β r_1 sin α=r_2 sin β⇒r_1 ^2 cos^2 α=r_1 ^2 −r_2 ^2 +r_2 ^2 cos^2 β c^2 −2cr_2 cos β+r_2 ^2 cos^2 β=r_1 ^2 −r_2 ^2 +r_2 ^2 cos^2 β cos β=((c^2 −r_1 ^2 +r_2 ^2 )/(2cr_2 )) ...(3) cos α=((c^2 −r_2 ^2 +r_1 ^2 )/(2cr_2 )) ...(4) Total overlapping area is X+Y as given in equation (1) and (2). value of α and β given in (3) and (4).

$$\mathrm{Common}\:\mathrm{chord}\:\mathrm{AB} \\ $$ $$\mathrm{center}\:\mathrm{O}_{\mathrm{1}} \:\mathrm{and}\:\mathrm{O}_{\mathrm{2}} \\ $$ $$\mathrm{O}_{\mathrm{1}} \mathrm{O}_{\mathrm{2}} \:\mathrm{cuts}\:\mathrm{AB}\:\mathrm{at}\:\mathrm{X} \\ $$ $$\angle\mathrm{AO}_{\mathrm{1}} \mathrm{X}=\alpha \\ $$ $$\angle\mathrm{AO}_{\mathrm{2}} \mathrm{X}=\beta \\ $$ $$\mathrm{AB}=\mathrm{2}{r}_{\mathrm{1}} \mathrm{sin}\:\:\alpha=\mathrm{2}{r}_{\mathrm{2}} \mathrm{sin}\:\beta \\ $$ $$\mathrm{O}_{\mathrm{1}} \mathrm{X}={r}_{\mathrm{1}} \mathrm{cos}\:\alpha,\:\:\:\:\mathrm{XO}_{\mathrm{2}} ={r}_{\mathrm{2}} \mathrm{cos}\:\beta \\ $$ $${r}_{\mathrm{1}} \mathrm{cos}\:\alpha+{r}_{\mathrm{2}} \mathrm{cos}\:\beta={c} \\ $$ $$\mathrm{area}\:\mathrm{of}\:\bigtriangleup\mathrm{AO}_{\mathrm{1}} \mathrm{B}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{AB}×\mathrm{O}_{\mathrm{1}} \mathrm{X}={r}_{\mathrm{1}} ^{\mathrm{2}} \mathrm{cos}\:\alpha\mathrm{sin}\:\alpha \\ $$ $$\mathrm{area}\:\mathrm{of}\:\bigtriangleup\mathrm{AO}_{\mathrm{2}} \mathrm{B}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{AB}×\mathrm{O}_{\mathrm{2}} \mathrm{X}={r}_{\mathrm{2}} ^{\mathrm{2}} \mathrm{cos}\:\beta\mathrm{sin}\:\beta \\ $$ $$\mathrm{area}\:\mathrm{of}\:\mathrm{overlapin}\:\mathrm{portion}=\mathrm{area}\:\mathrm{of} \\ $$ $$\mathrm{circle}\:\mathrm{segment}\:−\:\mathrm{area}\:\mathrm{of}\:\mathrm{triangle} \\ $$ $$\mathrm{Area}\:\mathrm{of}\:\mathrm{overlapping}\:\mathrm{portion}\:\mathrm{on}\:\mathrm{side}\:\mathrm{of} \\ $$ $$\mathrm{circle}\:\mathrm{with}\:\mathrm{radius}\:{r}_{\mathrm{1}} =\alpha{r}_{\mathrm{1}} −{area}\:\bigtriangleup\mathrm{AO}_{\mathrm{1}} \mathrm{B} \\ $$ $$\:\:\:\:\:\mathrm{X}=\frac{\mathrm{1}}{\mathrm{2}}\alpha{r}_{\mathrm{1}} ^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}{r}_{\mathrm{1}} ^{\mathrm{2}} \mathrm{sin}\:\mathrm{2}\alpha\:\:\:\:\:...\left(\mathrm{1}\right) \\ $$ $$\mathrm{Area}\:\mathrm{of}\:\mathrm{overlapping}\:\mathrm{portion}\:\mathrm{on}\:\mathrm{side}\:\mathrm{of} \\ $$ $$\mathrm{circle}\:\mathrm{with}\:\mathrm{radius}\:{r}_{\mathrm{2}} =\beta{r}_{\mathrm{2}} −{area}\:\bigtriangleup\mathrm{AO}_{\mathrm{2}} \mathrm{B} \\ $$ $$\:\:\:\:\:\mathrm{Y}=\frac{\mathrm{1}}{\mathrm{2}}\beta{r}_{\mathrm{2}} ^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}{r}_{\mathrm{2}} ^{\mathrm{2}} \mathrm{sin}\:\mathrm{2}\beta\:\:\:\:\:\:...\left(\mathrm{2}\right) \\ $$ $$\mathrm{Next}\:\mathrm{step}\:\mathrm{is}\:\mathrm{to}\:\mathrm{find}\:\alpha,\:\beta\: \\ $$ $${r}_{\mathrm{1}} \mathrm{cos}\:\alpha+{r}_{\mathrm{2}} \mathrm{cos}\:\beta={c}\Rightarrow{r}_{\mathrm{1}} ^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \alpha={c}^{\mathrm{2}} −\mathrm{2}{cr}_{\mathrm{2}} \mathrm{cos}\:\beta+{r}_{\mathrm{2}} ^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \beta \\ $$ $${r}_{\mathrm{1}} \mathrm{sin}\:\:\alpha={r}_{\mathrm{2}} \mathrm{sin}\:\beta\Rightarrow{r}_{\mathrm{1}} ^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \alpha={r}_{\mathrm{1}} ^{\mathrm{2}} −{r}_{\mathrm{2}} ^{\mathrm{2}} +{r}_{\mathrm{2}} ^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \beta \\ $$ $${c}^{\mathrm{2}} −\mathrm{2}{cr}_{\mathrm{2}} \mathrm{cos}\:\beta+{r}_{\mathrm{2}} ^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \beta={r}_{\mathrm{1}} ^{\mathrm{2}} −{r}_{\mathrm{2}} ^{\mathrm{2}} +{r}_{\mathrm{2}} ^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \beta \\ $$ $$\mathrm{cos}\:\beta=\frac{{c}^{\mathrm{2}} −{r}_{\mathrm{1}} ^{\mathrm{2}} +{r}_{\mathrm{2}} ^{\mathrm{2}} }{\mathrm{2}{cr}_{\mathrm{2}} }\:\:\:\:\:\:\:\:\:\:...\left(\mathrm{3}\right) \\ $$ $$\mathrm{cos}\:\alpha=\frac{{c}^{\mathrm{2}} −{r}_{\mathrm{2}} ^{\mathrm{2}} +{r}_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{2}{cr}_{\mathrm{2}} }\:\:\:\:\:\:\:\:\:...\left(\mathrm{4}\right) \\ $$ $$ \\ $$ $$\mathrm{Total}\:\mathrm{overlapping}\:\mathrm{area}\:\mathrm{is}\:\mathrm{X}+\mathrm{Y}\:\mathrm{as}\:\mathrm{given} \\ $$ $$\mathrm{in}\:\mathrm{equation}\:\left(\mathrm{1}\right)\:\mathrm{and}\:\left(\mathrm{2}\right).\:\mathrm{value}\:\mathrm{of}\:\alpha\:\mathrm{and}\:\beta \\ $$ $$\mathrm{given}\:\mathrm{in}\:\left(\mathrm{3}\right)\:\mathrm{and}\:\left(\mathrm{4}\right). \\ $$

Commented byYozzii last updated on 23/Dec/15

Isn′t the area of a sector of a circle of radius r produced by angle α given by (1/2)r^2 α? rα gives arc length no? Is the method different if both centres are on the same side of the commom chord?

$${Isn}'{t}\:{the}\:{area}\:{of}\:{a}\:{sector}\:{of}\:{a}\:{circle}\:{of} \\ $$ $${radius}\:{r}\:{produced}\:{by}\:{angle}\:\alpha\:{given}\:{by} \\ $$ $$\frac{\mathrm{1}}{\mathrm{2}}{r}^{\mathrm{2}} \alpha?\:\:\:{r}\alpha\:{gives}\:{arc}\:{length}\:{no}?\: \\ $$ $${Is}\:{the}\:{method}\:{different}\:{if}\:{both}\: \\ $$ $${centres}\:{are}\:{on}\:{the}\:{same}\:{side}\:{of}\:{the} \\ $$ $${commom}\:{chord}?\: \\ $$ $$ \\ $$

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