Intersection Area of two same large Circles

by Klaus-Anton, Jul 2, 2024, 7:45 AM

$[asy] //http://asymptote.ualberta.ca/ import geometry; size(8cm); unitsize(3cm); real r=1; real deg=60; pair O=(0,0) , A=(r,0) ,B=dir(deg) ,Bp=dir(-deg) ,C=dir(deg/2) ,Cp=conj(dir(deg/2)) ,D=(Cos(deg),0) ,myE=sqrt(3)/2*(Cos(deg/2),Sin(deg/2)) ,Ep=conj(myE) ; fill(arc(A,r,120,240)--cycle,lightgray); draw(circle(O,r)^^O--A--B--O--C); draw(circle(A,r)); draw(O--Bp--A^^O--Cp); draw(O--O-1,dashed); draw(B--Bp,Dotted+gray); dot(O^^A^^B^^Bp^^C^^Cp^^D^^myE^^Ep,UnFill); perpfactor=.8; perpendicularmark((.5*r,0),NE); perpendicularmark(myE,dir(myE--O)*NE); perpendicularmark(Ep,dir(Ep--O)*SE); label(format("$r=%3f$",r),O--O-1); label("$O$",O,SW*dir(3)); label("$A$",A,.85*E); label("$B$",B,N*dir(-2)); label("$B^\prime$",Bp,1.1*S*dir(-4)); label("$C$",C,.85dir(deg/2)); label("$C^\prime$",Cp,.65dir(-deg/2)*dir(5)); label("$D$",D,.9*SE*dir(-6)); label("$E$",myE,1.375*dir(C--O,A--B)); label("$E^\prime$", Ep,1.1*dir(A--Bp,Cp--O)*dir(8)); shipout(bbox(2mm,FillDraw(white,white))); [/asy]$

Figure 1

Figure 2: The Real $\large{\textit{\textbf{r}}}$ can be modified. - And everything else still fits.

$[asy] //http://asymptote.ualberta.ca/ import geometry; size(8cm); unitsize(3cm); real r=1.5; real deg=60; pair O=(0,0) , A=(r,0) ,B=r*dir(deg) ,Bp=r*dir(-deg) ,C=r*dir(deg/2) ,Cp=conj(C) ,D=r*(Cos(deg),0) ,myE=r*sqrt(3)/2*(Cos(deg/2),Sin(deg/2)) ,Ep=conj(myE) ; fill(arc(A,r,120,240)--cycle,lightgray); draw(A--A+r,lightgray); draw(line(O,false,B),lightgray); draw(line(O,Bp),lightgray); draw(line(O,C),lightgray); draw(line(O,Cp),lightgray); draw(line(A,B),lightgray); draw(line(A,Bp),lightgray); draw(line(B,Bp),lightgray); draw(line(myE,false,shift(myE)*dir(myE--O,myE--B)),lightgray); draw(line(Ep,false,shift(Ep)*dir(Ep--O,Ep--Bp)),lightgray); draw(circle(O,r)^^O--A--B--O--C); draw(circle(A,r)); draw(O--Bp--A^^O--Cp); draw(O--O-r,dashed); draw(B--Bp,Dotted+gray); draw(circle(O,1/10*r),gray); draw(circle(A,1/10*r),gray); draw(circle(B,1/10*r),gray); draw(circle(C,1/10*r),gray); draw(circle(D,1/10*r),gray); draw(circle(Bp,1/10*r),gray); draw(circle(Cp,1/10*r),gray); draw(circle(D,1/10*r),gray); draw(circle(myE,1/10*r),gray); draw(circle(Ep,1/10*r),gray); draw(box((-1.1*r,-1.1*r),(1.1*r+r,1.1*r)),invisible); dot(O^^A^^B^^Bp^^C^^Cp^^D^^myE^^Ep,UnFill); perpfactor=.8; perpendicularmark((.5*r,0),NE); perpendicularmark(myE,dir(myE--O)*NE); perpendicularmark(Ep,dir(Ep--O)*SE); label(format("$r=%3f$",r),O--O-r); label("$O$",O,SW*dir(3)); label("$A$",A,.85*E); label("$B$",B,N*dir(-2)); label("$B^\prime$",Bp,1.1*S*dir(-4)); label("$C$",C,.85dir(deg/2)); label("$C^\prime$",Cp,.65dir(-deg/2)*dir(5)); label("$D$",D,.9*SE*dir(-6)); label("$E$",myE,1.375*dir(C--O,A--B)*dir(3)); label("$E^\prime$", Ep,1.1*dir(A--Bp,Cp--O)*dir(8)); shipout(bbox(2mm,FillDraw(white,white))); [/asy]$

Figure 3: How good is the Fine-Tuning of the Labels? - The Letter ${\textit{\textbf{E}}}$ was rotated a little bit.

Okay. So far so good. We now have put enough work into the only formal apprearance of the graphic.

Let's go! Now: In medias res. The actual task is to find the area of here - have a look - in lightred marked partial figure. - Hm.

$[asy] import graph; real r=1; real deg=60; real GR=(sqrt(5)+1)/2; real gr=(sqrt(5)-1)/2; pair O=(0,0),A=(r,0),B=dir(deg),Bp=conj(B),C=(Cos(deg),0),D=midpoint(A--B); draw(circle(O,r)); /* filldraw(Arc(O,Bp,B,n=32000)&B--Bp--cycle,lightred+white,red+linewidth(GR));//runtime: non-cyclic path cannot be filled fill(f,t*g,p,false); */ filldraw(arc(O,Bp,B)--cycle,lightred+white,red+linewidth(GR)); draw(O--A--B--O--Bp--A^^O--D); label("$O$",O,GR^.5*W); label("$A$",A,E); label("$B$",B,dir(B)); label("$B^\prime$",Bp,dir(Bp)); label("$C$",C,(1+gr^4)*SW); label("$D$",D,GR*dir(D--O,D--A)); shipout(bbox(2mm,FillDraw(white,white))); [/asy]$

Figure 4: Simplified Drawing, Circle Segment

$[asy] import graph; real r=1; real deg=60; real GR=(sqrt(5)+1)/2; real gr=(sqrt(5)-1)/2; pair O=(0,0),A=(r,0),B=dir(deg),Bp=conj(B),C=(Cos(deg),0),D=midpoint(A--B); draw(circle(O,r)); /* filldraw(Arc(O,Bp,B,n=32000)&B--Bp--cycle,lightred+white,red+linewidth(GR));//runtime: non-cyclic path cannot be filled fill(f,t*g,p,false); */ fill(arc(O,Bp,B)--O--cycle,lightblue+white); draw(O--A--B--Bp--A^^O--D); Draw(arc(O,Bp,B)--O--cycle,blue+linewidth(GR)); label("$O$",O,GR^.5*W); label("$A$",A,E); label("$B$",B,dir(B)); label("$B^\prime$",Bp,dir(Bp)); label("$C$",C,(1+gr^4)*SW); label("$D$",D,GR*dir(D--O,D--A)); shipout(bbox(2mm,FillDraw(white,white))); [/asy]$

Figure 5: Circle Sector

Aha! - Slowly i think it is coming me. Let us throw away the segment $OD$ , and so we will also drop the point $D$ . Okay.

$[asy] import graph; real r=1; real deg=60; real GR=(sqrt(5)+1)/2; real gr=(sqrt(5)-1)/2; pair O=(0,0),A=(r,0),B=dir(deg),Bp=conj(B),C=(Cos(deg),0); fill(arc(O,Bp,B)--O--cycle,lightblue+white); draw(circle(O,r)^^O--A--B--Bp--A); draw(arc(O,Bp,B)--O--cycle,blue+linewidth(GR)); label("$O$",O,GR^.5*W); label("$A$",A,E); label("$B$",B,dir(B)); label("$B^\prime$",Bp,dir(Bp)); label("$C$",C,(1+gr^4)*SW); shipout(bbox(2mm,FillDraw(white,white))); [/asy]$

Figure 6: Circle Sector - More Simplification

Oh yes. I have more and more the impression that i am on the best way that i soon will have got it. Yes, yes, yes. Thank you. Oh thank you very much! Thym.

This is fine. Look! This circle sector will repaet in the circle three times. If i can trust my eyes. Let us go out from the situation that the arc from $A$ to $B$ about $O$ - the arc $\overarc{AOB}$ , that this arc is: $\overarc{AOB}=60^\circ$ .

Hm. Okay.

$[asy] import graph; real r=1; real deg=60; real GR=(sqrt(5)+1)/2; real gr=(sqrt(5)-1)/2; pair O=(0,0),A=(r,0),B=dir(deg),Bp=conj(B),C=(Cos(deg),0); fill(arc(O,Bp,B)--O--cycle,lightblue+white); draw(O--dir(180),Dotted+blue); draw(circle(O,r)^^O--A--B--Bp--A); draw(arc(O,Bp,B)--O--cycle,blue+linewidth(GR)); label("$O$",O,1.1*dir(240)*dir(-3)); label("$A$",A,E); label("$B$",B,dir(B)); label("$B^\prime$",Bp,dir(Bp)); label("$C$",C,(1+gr^4)*SW); shipout(bbox(2mm,FillDraw(white,white))); [/asy]$

Figure 7: Circle Sector - Belonging to the Arc of the $\angle B^\prime \kern-.75ptOB$ - Threefold Repitition

This what i can imagine me now that is, that the area belonging to this arc $\angle B^\prime \kern-.75ptOB$ is one third of the area of the whole circle.

Yes! - This it is. Oh yeah, yeah. And now?

Now it is so easy. From the circle sector you only will have to substract the area of the triangle $\Delta \kern.5ptOB^\prime \kern-1.5ptB$ . So, so simple!

https://upload.wikimedia.org/wikipedia/commons/thumb/b/b2/CIRCLE_LINES-en.svg/238px-CIRCLE_LINES-en.svg.png

https://upload.wikimedia.org/wikipedia/commons/thumb/1/12/Circle_slices-en.svg/240px-Circle_slices-en.svg.png

Figure 8a and Figure 8b

(Direct Linkls to wiki-Circle)

And so you also have the right outlook for the correct way to find the intersection area of two circles. You remember. This was our outgoing problem (compare and see there to on Figure 1).

$[asy] //http://asymptote.ualberta.ca/ import geometry; size(8cm); unitsize(3cm); real r=1; real deg=60; pair O=(0,0) , A=(r,0) ,B=dir(deg) ,Bp=dir(-deg) ,C=dir(deg/2) ,Cp=conj(dir(deg/2)) ,D=(Cos(deg),0) ,myE=sqrt(3)/2*(Cos(deg/2),Sin(deg/2)) ,Ep=conj(myE) ; fill(arc(A,r,120,240)--cycle,lightgray); import patterns; real gr=(sqrt(5)-1)/2; add("hatch",hatch(gr^-1*mm,blue)); filldraw(arc(A,r,120,240)--cycle,pattern("hatch")); fill(arc(O,r,-60,60)--cycle,lightgray); draw(circle(O,r)^^O--A--B--O--C); draw(circle(A,r)); draw(O--Bp--A^^O--Cp); draw(O--O-1,dashed); draw(B--Bp,Dotted+gray); dot(O^^A^^B^^Bp^^C^^Cp^^D^^myE^^Ep,UnFill); perpfactor=.8; perpendicularmark((.5*r,0),NE); perpendicularmark(myE,dir(myE--O)*NE); perpendicularmark(Ep,dir(Ep--O)*SE); label(format("$r=%3f$",r),O--O-1); label("$O$",O,SW*dir(3)); label("$A$",A,.85*E); label("$B$",B,N*dir(-2)); label("$B^\prime$",Bp,1.1*S*dir(-4)); label("$C$",C,.85dir(deg/2)); label("$C^\prime$",Cp,.65dir(-deg/2)*dir(5)); label("$D$",D,.9*SE*dir(-6)); label("$E$",myE,1.375*dir(C--O,A--B)); label("$E^\prime$", Ep,1.1*dir(A--Bp,Cp--O)*dir(8)); shipout(bbox(2mm,FillDraw(white,white))); [/asy]$

Figure 9

Now finally, i think we really have enough. We can begin to calculate.

What is the area of the circle with radius $r=1$ ? - wiki-Circle explains us:

$\textrm{Area}\ =\ \pi\times r^2.$
Equivalently - denoting the diameter by $d$ - we have:

$\textrm{Area}\ =\ \frac{\pi\times d^2}{4}\approx0.785\,398\times d^2.$
And so the area of the circle with radius $r=1$ - the so called unitcircle - is exactly:

$\textrm{Area}\ =\ \pi\times r^2=\pi\times1^2=\pi=3.141\,592\dots$
So the Circle Sector - belonging to the arc of the $\angle B^\prime \kern-.75ptOB$ has as area:
$\pi/3=3.141\,59\dots/3=1.047\,195\dots$
But with this intermediate result we are not at all arrived at the end. In the next step we need the area of the corresponding Circle Segment.

Here, please take breath. Two times. - Three times. Make a sigher. Make a sigh of relief. A deep profund sigh. And as i know you love Maths - make an amourous sigh. Pausing - $\texttt{\textbackslash relax}$ as says. ( $\textit{"Do nothing!"}$ , ${\textit{The}\ T\kern -.41667em\lower .5ex \hbox{\bf \it E}\kern -.125em X}\kern-1ptbook$ , page 34.)

And. What do Mathematicians like? Have a look into the ${T\kern -.41667em\lower .5ex \hbox{\bf \it E}\kern -.125em X}\kern-1ptbook$ (texbook.pdf of D.E. Knuth and search for the keyword "like").

$[asy] //unitsize(2.5cm);//viel zu groß // aber: invisible brace corrects?//nein //size(8cm);// So passt es zur invisible brace. size(6cm); import geometry; pair O,A,B,C,Cp; real r=1 ,GR=(sqrt(5)+1)/2 ,dirA=0 ,dirB=120 ,dirC=60; pair O=(0,0) ,A=dir(dirA) ,B=dir(dirB) ,C=dir(dirC) ,Cp=midpoint(A--B); label("$\underbrace{\makebox[2.5cm]{\phantom{}}}$",(.5,-.075),invisible); //fill(arc(O,A,B)--O--A,lightblue);// runtime: non-cyclic path cannot be filled fill(f,t*g,p,false); //fill(arc(O,A,B)--O--cycle,lightblue);// works (sector) filldraw(arc(O,A,B)--cycle,lightblue,.8blue+linewidth(GR));// works (segment) draw(//circle(O,r) arc(O,B,A) ^^O--A ^^O--B ^^O--Cp ); draw(Cp--C,white+dotted); label("$O$",O,dir(210)); label("$A$",A,dir(dirA),.8blue); label("$B$",B,dir(dirB),.8blue); label("$C$",C,dir(dirC),.8blue); label("$C^\prime$",Cp,1.1*dir(dirC+90+45+8),.8blue); label("$1$",O--A); label("$1$",B--O); dot(O,UnFill); dot(A^^B^^C^^Cp,blue,UnFill); shipout(bbox(2mm, FillDraw(white,white))); [/asy]$

Figure 10a: What is the Area of the lightblue Circle Segment AB

$[asy] // Figure 10b unitsize(2.5cm); import geometry; pair O,A,B,C,Cp; real r=1 ,GR=(sqrt(5)+1)/2 ,dirA=0 ,dirB=120 ,dirC=60; pair O=(0,0) ,A=dir(dirA) ,B=dir(dirB) ,C=dir(dirC) ,Cp=midpoint(A--B); //fill(arc(O,A,B)--O--A,lightblue);// runtime: non-cyclic path cannot be filled fill(f,t*g,p,false); //fill(arc(O,A,B)--O--cycle,lightblue);// works (sector) fill(arc(O,A,B)--cycle,lightblue);// works (segment) perpfactor=.8; perpendicularmark(Cp,dir(B--A)*NE,white); draw(arc(O,A,B)--cycle,.8blue+linewidth(GR));// works (segment) draw(arc(A,O,B,.5),blue); draw(//circle(O,r) arc(O,B,A) //^^O--A--B--cycle ^^O--A ^^O--B ^^O--Cp ); draw(Cp--C,white+dotted); label("$O$",O,1.5*dir(240)); label("$A$",A,dir(dirA),.8blue); label("$B$",B,dir(dirB),.8blue); label("$C$",C,dir(dirC),.8blue); label("$C^\prime$",Cp,1.1*dir(dirC+90+45+8),.8blue); label("$1$",O--A,2.75S); label("$\underbrace{\makebox[2.5cm]{\phantom{}}}$",(.5,-.075)); label(rotate(-60)*"$\underbrace{\makebox[2.5cm]{\phantom{}}}$",midpoint(O--B),.5*dir(dir(120+90))); label("$1$",B--O,2.5*dir(O--B)*I); label("$0.5$",(.25,0),N); dot(O,UnFill); dot(A^^B^^C^^Cp,blue,UnFill); //dot(midpoint(O--C)); shipout(bbox(2mm, FillDraw(white,white))); [/asy]$

Figure 10b: $\overarc{AOB}$ $\ =120^\circ$

You can calculate the right triangle $\Delta(O,A,C^\prime)$ with the help of Pythagoras. Let us say $OC^\prime=a=1$ is the one leg and $AC^\prime=b$ the other leg, from which you searching the length, and $OA=1=c$ is the hypothenuse. Go with your known values for example to Mathpower Pythagoras-Rechner and you will get the solution including the solution way. We find there $b=8.666\ldots=\sqrt3/2$ . And they tell us there at the end that this triangle has an area of $\Delta(O,A,C^\prime)_\textrm{A}=.217\dots$

$\left[p=a^2/c=.5^2/1\Longrightarrow q=c-p=1-.25=.75\right]$

$\left[ h=\sqrt{p\times q}=\sqrt{.25\times.75}=.25\sqrt3=.433\dots\right]$

$\left[ \Delta(O,A,C^\prime)_\textrm{A}= \frac{c\times h}{2}= \frac{1\times \sqrt{.25\times.75}} {2}=.125\sqrt3=.217\dots \right]$

$[asy] //Figure 11 unitsize(2.5cm); import geometry; pair O,A,B,C,Cp; real r=1 ,GR=(sqrt(5)+1)/2 ,dirA=0 ,dirB=120 ,dirC=60; pair O=(0,0) ,A=dir(dirA) ,B=dir(dirB) ,C=dir(dirC) ,Cp=midpoint(A--B); //fill(arc(O,A,B)--O--A,lightblue);// runtime: non-cyclic path cannot be filled fill(f,t*g,p,false); //fill(arc(O,A,B)--O--cycle,lightblue);// works (sector) //fill(arc(O,A,B)--cycle,lightblue);// works (segment) //perpfactor=.8; //perpendicularmark(Cp,dir(B--A)*NE,white); //draw("$120^\circ$",arc(O,.25,0,120),.8blue);//autosizes filldraw(arc(O,A,B)--O--A--cycle,lightblue,.8blue+linewidth(GR));// works (segment) //draw(arc(O,.2,0,120),.8blue); //arc to 117 degree and not to 120 degree, because as correction to PenMargin, asy it meats there a thicker line //draw(arc(O,.2,0,117),.8blue,Arrow(size=2mm),PenMargin); //instead: PenMargins takes two optional arguments. // see: https://tex.stackexchange.com/questions/526257/what-is-the-problem-of-markers2-asy-for-texhead/526351#526351 draw(arc(O,.2,0,120),.8blue,Arrow(size=2mm),PenMargin(0,GR)); //draw(arc(A,O,B,.5),blue); draw(//circle(O,r) arc(O,B,A) //^^O--A--B--cycle //^^O--A //^^O--B //^^O--Cp ); //draw(Cp--C,white+dotted); // Not on the midpoint. - This seems so nicer for the eye as i think. label(rotate(-30)*"$120^\circ$",relpoint(arc(O,.2,0,120), .4),dir(C),.8blue); label(rotate(-30)*"$\textrm{A}=\pi/3$",Cp,.3*dir(O--C),white); label("$O$",O,1.5*dir(240),.8blue); label("$A$",A,dir(dirA),.8blue); label("$B$",B,dir(dirB),.8blue); //label("$C$",C,dir(dirC),.8blue); //label("$C^\prime$",Cp,1.1*dir(dirC+90+45+8),.8blue); label("$1$",O--A,2.75S); label("$\underbrace{\makebox[2.5cm]{\phantom{}}}$",(.5,-.075)); label(rotate(-60)*"$\underbrace{\makebox[2.5cm]{\phantom{}}}$",midpoint(O--B),.5*dir(dir(120+90))); label("$1$",B--O,2.5*dir(O--B)*I); //label("$0.5$",(.25,0),N); //dot(O,UnFill); dot(O^^A^^B,blue,UnFill); //dot(relpoint(O--C,.5)); shipout(bbox(2mm, FillDraw(white,white))); [/asy]$

Figure 11: Area of Circle Sector

$\textrm{A}=\frac{\pi}{3}=1.047\,198\dots$

AoPS wiki Sector says:

AoPS wiki Sector wrote:

[...]
The area of a sector is found by multiplying the area of circle $O$ by $\frac{\theta}{2\pi}$ , where $\theta$ is the central angle in radians.

Therefore, the area of a sector is $\frac{r^2\theta}{2}$ , where $r$ is the radius and $\theta$ is the central angle in radians.

Alternatively, if $\theta$ is in degrees, the area is $\frac{\pi r^2\theta}{360^{\circ}}$ .

$[asy] //Figure 12 unitsize(2.5cm); import geometry; pair O,A,B,C,Cp; real r=1 ,GR=(sqrt(5)+1)/2 ,dirA=0 ,dirB=120 ,dirC=60; pair O=(0,0) ,A=dir(dirA) ,B=dir(dirB) ,C=dir(dirC) ,Cp=midpoint(A--B); //fill(arc(O,A,B)--O--A,lightblue);// runtime: non-cyclic path cannot be filled fill(f,t*g,p,false); //fill(arc(O,A,B)--O--cycle,lightblue);// works (sector) //fill(arc(O,A,B)--cycle,lightblue);// works (segment) //perpfactor=.8; //perpendicularmark(Cp,dir(B--A)*NE,white); //draw("$120^\circ$",arc(O,.25,0,120),.8blue);//autosizes filldraw(arc(O,A,B)--cycle,lightblue,.8blue+linewidth(GR));// works (segment) draw(arc(O,.2,0,120),Arrow(size=2mm),PenMargin); //draw(arc(A,O,B,.5),blue); draw(//circle(O,r) arc(O,B,A) //^^O--A--B--cycle ^^O--A ^^O--B //^^O--Cp ); //draw(Cp--C,white+dotted); // Not on the midpoint. So it seems me to be nicer. label(rotate(-30)*"$120^\circ$",relpoint(arc(O,.2,0,120),.4),.75*dir(C)); //label(scale(.85)*rotate(-30)*"$\textrm{A}=\pi/3-\sqrt{.25\times.75}$",relpoint(O--C,.5),.5*dir(O--C),white); label(scale(.85)*rotate(-30)*"$\textrm{A}=\pi/3-.25\sqrt{3}$",relpoint(O--C,.55),.5*dir(O--C),white); label("$O$",O,1.5*dir(240)); label("$A$",A,dir(dirA)*dir(-30/2),.8blue);// -dir(A--B,A--O) label("$B$",B,dir(dirB)*dir(30/2),.8blue);// -dir(B--A,B--O) //label("$C$",C,dir(dirC),.8blue); //label("$C^\prime$",Cp,1.1*dir(dirC+90+45+8),.8blue); label("$1$",O--A,2.75S); label("$\underbrace{\makebox[2.5cm]{\phantom{}}}$",(.5,-.075)); label(rotate(-60)*"$\underbrace{\makebox[2.5cm]{\phantom{}}}$",midpoint(O--B),.5*dir(dir(120+90))); label("$1$",B--O,2.5*dir(O--B)*I); //label("$0.5$",(.25,0),N); distance(Label("$\sqrt3$",RightSide),A,B,1.5cm,joinpen=grey+dashed); dot(O,UnFill); dot(A^^B,blue,UnFill); shipout(bbox(2mm, FillDraw(white,white))); [/asy]$

Figure 12: Area of Circle Segment

$\textrm{A}=\frac{\pi}{3}-.25\sqrt3=0.614\,185\dots$

$\left[\sqrt{.25\times.75}=\kern-1pt.25\sqrt3\right]$

What do i can say to you? How to calculate the area of a triangle. The easiest way is:

$\textrm{A}=\frac{\textrm{groundside}\times \textrm{height}}{2}=\frac{\textrm{base}\times\textrm{height}}{2}$
And now, here concrete - okay.

$\textrm{A}_{\Delta(ABC)}=\frac{\sqrt3\times .5}{2}=\sqrt3/4\approx0.433\,012$
Finally we can say that the intersection area of two circles with radii $r=1$ intersecting at $\pm 60^\circ$ is twice the segment of the circle with this size:

$\textrm{A}_{(r=1)}=\frac{8\pi-6\sqrt3}{12}=\frac{2\pi}{3}-\frac{\sqrt3}{2}=1.228\,369\dots$
And for two circles with radii $r=1.5$ :

$\textrm{A}_{(r=1.5)}=1.5^2\times \left(\frac{2\pi}{3}- \frac{\sqrt3}{2}\right)=2.763\,831\dots$
What we here have found out is covered with that what here is said on page 2: A. Kratochwill 2016 K06.pdf / Kreis - Schnittflächen.

Notation Note:
As i am having here a point which i called $A$ i have made the (incorrect) typographically distinguishing decision to write the area down in text roman as $\textrm{A}$ . I agree so far that this was an incorrect decision, because upright notation in Maths is (mostly / normally) reserved for constants.

wiki wrote:

Later, René Descartes (17th century) introduced the modern notation for variables and equations; in particular, the use of $x,y,z$ for unknown quantities and $a,b,c$ for known ones (constants). He introduced also the notation $i$ and the term "imaginary" for the imaginary unit.