I have an equation to calculate the distance to the outside of a circle from an eccentric point within the circle.
$$x = E\cos(a) + 0.5\sqrt{(D^2) - 4*(E^2)\sin(a)^2}$$
Where: $E$ = eccentricity, $D$ = Circle diameter, $a$ = angle (the maximum distance between the eccentric point and the circle perimeter is at 0 degrees)
If the centre point of the circle is labelled $A$, the eccentric point is labelled $B$ and the point on the perimeter of the the circle at any given angle is $C$, the equation gives the length of line $BC$ at a a given angle between line $BC$ and line $BA$.
I thought there would be a way to integrate this equation to get an area thats is bounded by lines at two different angles and the perimeter of the circle, but my calculus and trigonometry skills are not up to scratch.
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$\begingroup$Symbols slightly altered.. $( \rho, R, \theta, \epsilon ) = ( - , D/2, \alpha, E)$ correspond.
Eccentric radius $\rho$, Circle radius $R$, Eccentric polar angle $\theta $
$$ \epsilon^2 + \rho^2 - 2 \epsilon \rho \cos \theta = R^2 \tag{1}$$
Solve quadratic and taking positive sign to evaluate to right of eccentricity,
$$ \rho = \epsilon \cos \theta + \sqrt{R^2 - ( \epsilon \sin \theta)^2 ) } \tag{2} $$
Its square is needed in area polar formula
$$ \rho^2 = \epsilon ^2 \cos ( 2 \theta) + R^2 +( 2 \epsilon \cos \theta) \sqrt{R^2 - ( \epsilon \sin \theta)^2 } \tag {3} $$
$$ 2\, dArea = \rho^2 d \theta \, \tag{4}$$
Integrate within limits w.r.t. $\theta $ independent variable.
$$ 2 Area = \int_{\theta_1} ^{\theta_2} \rho^2 d \theta \, \tag{5} $$
Hope I made no error and hope you will take it to finish the integral evaluation.
A closed form solution got by Mathematica...
$\endgroup$ $\begingroup$Just add the area of the sector centered at $A$ to that of the triangle $CBA$.
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