Curved Lines in a Circle

* If angle x is in radian, curved side BKC=ax & curved side DKE=ax1. ..........(1)

* BC=2a*sin(x/2) and DE=2a*sin(x1/2) .....(1a)                          Since BC2=a2+a2-2a2cos x

* BkC/BC=(x/2)/sin (x/2)  and DkE/DE=(x1/2)/sin (x1/2) .........(1b)

    As x -> 0, the RHS tends to 1 and Bkc=BC and DkE=DE

* AL/KL= a*cos (x/2)/[a*(1-cos (x/2)]=1/[1/cos(x/2)  -1]  and AT/KT= 1/[1/cos(x1/2)  -1] ......(1c)

* Perimeter of the curved triangle ABkC is a(2+x) & curved triangle ADkE=a(2+x1) ........(2)

* Area of the curved triangle ABkC is (π*a2/2π)*x = a2x/2=(1/2)*ax*a .........(3)

* Area of the curved triangle ADkE is (π*a2/2π)*x1 = a2x1/2=(1/2)*ax1*a ........(4)

* Analogous to a triangle ABC where a/(b+c-a) +b/(a+c-b)+c/(a+b-c) =z which has a minimum value 3, here

   z=2/ x  +x/(2-x)  for Curved triangle ABkC ........(5)

  z1=2/ x1  +x1/(2-x1)  for Curved triangle ADkE ......(6)

* Triangular inequality : a+ax >a  => ax > 0 which is true  .....(7)

* Triangular inequality : a+a >ax  may or may not hold good unlike a st. line triangle where it invariably holds good.

* Putting the cosine law, a2x2=a2+a2-2a2cos Y where Y is a hypothetical angle, we get Y=2*sin-1(x/2)

* Similarly, Putting the cosine law, a2=a2x2+a2-2x2a2cos U where U is a hypothetical angle, we get U=cos-1(x/2)

* In curved triangle ABkC, sum of angles Y+2U=  2*sin-1(x/2)  +2*cos-1(x/2)    = 2*π/2 = π where Y,U are hypothetical angles. Same for the other curved triangle.

While analysing the equations of curved triangles, one finds that the area formula is equivalent to that of straight triangle.The triangular inequality partly holds good. Cosine formula can be fitted in with hypothetical angles etc.

Explanation: Suppose slope of AE is m2, Co-ordinate of E is (x2,y2), then m2=(y2-h1)/(x2-g1) or y2=m2x2+(h1-g1m2)=m2x2+k where k=h1-g1m2 or y2=m2x2+k.......(5)

Now, (g1-x2)2 +(h1-y2)2  =a2  or (g1-x2)2 +(h1-k-m2x2)2  =a2  or  (g1-x2)2 +(g1m2-m2x2)2  =a2  or on simplification, we get                      (1+m22 )x22  + (-2g1)(1+m22 )x2 + (g12m22 +g12 - a2 ) =0  which is a quadratic equn. which gives 2 value of x2 for each value of m and hence we get in total 4 values of x2.

* If the slope of AE is m2', then the x & y co-ordinate of E (m2,n2) are related by formula m2m2' -g1m2'+h1=n2 ; and m2 can be found out by solving the quadratic equation (1+m2'*m2')m2*m2 +2g1(m2'*m2' -1)m2 +(g1*g1+g1*g1*m2'*m2'-a*a)=0. since m2' has 2 values and for each value of m2', m2 has 2 values => m2 has total 4 values .

* TZ=a*sin(x1r/2)*tan(x1r/2) and AZ=AT+TZ=a / Cos(x1r/2). Previously, TZ was written as [(a*cosx1)/(2*cos(x1/2))] which is wrong. pl. check.

* Value of AT, TZ found out from length and from co-ordinates tally for angle > 0 and up to 162.54 degree and at 270 degree. From co-ordinates , we take AT1 and AT8 for AT and TZ1 & TZ4 for TZ. Reasons not yet explored.

Value of a
Value of x in Degree
Value of x1 in Degree
Data below is independent of data above  
x-coordinate of center at A(g1)
y-coordinate of center at A(h1)
x-coordinate of D-put any arbitrary value (m1)
x in radian
Comment on x
x1 in radian
Comment on x1
DZ=ZE [a1=a/(2*cos (x1/2))]
a1/a  [1/(2*cos (x1/2)]
AT (a*cos[x1/2])
TZ  [a*sin(x1/2)*tan(x1/2)]
AZ [a/( cos (x1/2)]
area of Δ DAE = a2 * sinx1 / 2
area of Δ DZE = a12 * sin(180°-x1) / 2 =(a12 *sinx1 )/ 2
area DAEZ =(  a2+a12 )*(sinx1) / 2
area of Δ DAE / area of Δ DZE
BkC arc
DkE arc
area of DBkCZ (red area)=[sinx1*(a2+a12) /2] - a2x1/2
sum of 2 sides/3rd side(2a/ax) in ABkC
sum of 2 sides/3rd side(2a/ax1) in ADkE
Perimeter of ABkC:P1 [a(2+x)]
Perimeter of ADkE:P2 [a(2+x1)]
Area of ABkC:A1
Area of ADkE:A2
Area of DkET:A3 [(a2/2)*(x1-sinx1)]
Area of BkCL:A4 [(a2/2)*(x-sinx)]
A3/A4 [(x1-sinx1)/(x-sinx)], x,x1 in radian
Angle Y
Angle U
Y+2U in degree
Angle Y1
Angle U1
Y1+2U1 in degree
If BC becomes the new arc, corresponding angle x2 is (radian) AND (degree)
If BC becomes the new arc of the circle, % change in angle x is(-Ve) % (this is independent of magnitude of radius)
If angle between AB & AC are reduced to zero, maximum stretch of chord beyond k is DE/2 = asin(x1/2) or BC/2=a*sin(x/2) as the case may be. Taking start point as BC, stretch is asin(x/2)
Look to fig1 and fig. 2 and find perimeter , area of fig 2. Green lines are tangent to the circle with radius a and join at the center of circle
Perimeter (perimeterfactor*a)
Area         (areafactor*a2)
Area factor
Perimeter factor
General Converter    
Put angle in Radians
Find angle in Degree
Equation of the circle (x-  )2 +(y-)2 = =
or simplified equation of circle x2 + y-  () x - () y + () = 0
y-coordinate of D-(n1a) & (n1b) =± √[a2-(m1-g1)2] +h1   and 
Slope of AD w.r.t.  x-axis   and 
Slope of DZ w.r.t.  x-axis (DZ is perpendicular to AD)   and 
Y-Intercept of DZ   ,  ,,
Equn. of st. line DZ......(1) y=x +