There is a circle with radius r1 and centre D
This intersects a circle with radius r2 and centre C
Tangent line AB is always tangential to circle with centre D
It can be assumed the circle with centre C is free to move, and can be both inside and outside of circle D, with centre C also being able to be either side of the tangent line - I will update with edited images and scenarios (an all encompassing is the dream haha!)
Updated query - Is there a way to find the area bound shaded green/yellow? I have been around the houses subtracting sectors and am struggling to find my mistakes, to the point I believe my apporach may well be (is!) off, so I'm not looking to lead anyone!
UPDATE - That said, hold the phone, can we get there by adding green (just intersecting circles equation), and yellow as being:
Yellow = [Area_tri BPA] - [Area_Segment BDP] + [Area_Segment ACP]
I updated the image to better reflect all scenarios of where circle C can be in relation to the tangent line and circle D
I'll clean this query all up once we resolve so it is less cluttered!
I have code which currently can move circle C so that it intersects only the tangent line (area cut from C = segment) and also for when it is only the circles C&D intersecting (area cut from C = circles intersection) and when no intersection with tangent or circle D, (area cut from C = 0)
Let the point $D$ of your diagram be the origin of the coordinate system and define the center of the smaller circle at point $C$ $(x_C,y_C)$. We then have the equations for the two circles (when $y_C < 0$), $$ y=\pm \sqrt{r_1^2 -x^2}\\ y=\pm \sqrt{r_2^2 -(x+x_C)^2}-y_C $$ To find the intersection of the two circles, points $P$ and $Q$, equate the above equations to determine $(x_P,y_P)$ and $(x_Q,y_Q)$. Doing the algebra we find, $$ x_P = -x_C (\frac{ x_C^2 + r_1^2 - r_2^2}{x_C^2 + y_C^2}) + \sqrt{x_C^2 (\frac{ x_C^2 + r_1^2 - r_2^2}{x_C^2 + y_C^2})^2-\frac{[(x_C^2 - y_C^2 + r_1^2 - r_2^2)^2 + 4y_C^2 (x_C^2 - y_C^2)]}{x_C^2 + y_C^2}}\\ y_P=\sqrt{r_1^2 - x_P^2}\\ x_Q = -x_C (\frac{ x_C^2 + r_1^2 - r_2^2}{x_C^2 + y_C^2}) - \sqrt{x_C^2 (\frac{ x_C^2 + r_1^2 - r_2^2}{x_C^2 + y_C^2})^2-\frac{[(x_C^2 - y_C^2 + r_1^2 - r_2^2)^2 + 4y_C^2 (x_C^2 - y_C^2)]}{x_C^2 + y_C^2}}\\ y_Q=-\sqrt{r_1^2 - x_Q^2} $$
Denote the line segment $PQ$ by $L$ $$ L^2=(x_P - x_Q)^2 + (y_P - y_Q)^2 $$ and from the opposite angle formula of trigonometry, $$ L^2= 2r_1^2 - 2r_1^2\cos(\theta_1)\\ \cos(\theta_1)=1-\frac{L^2}{2r_1^2} $$ The area of the isosceles triangle in the big circle is $$ \frac{1}{2}Lh\\ h=r_1\cos(\frac{\theta_1}{2})\\ \cos(\frac{\theta_1}{2})=\sqrt{\frac{\cos(\theta_1)+1}{2}}=\sqrt{1-\frac{L^2}{4r_1^2}}\\ \frac{1}{2}Lh=\frac{1}{2}Lr_1 \sqrt{1-\frac{L^2}{4r_1^2}} $$ The area of the piece of the pie subtended in the big circle by $\theta_1$ is $\frac{1}{2}r_1^2 \theta_1$, therefore the area of overlap below the line segment $L$ is $$ \frac{1}{2}r_1^2\cos^{-1}(1-\frac{L^2}{2r_1^2})-\frac{1}{2}Lr_1 \sqrt{1-\frac{L^2}{4r_1^2}} $$ In like manner we find the area of overlap above the line segment $L$, $$ \frac{1}{2}r_2^2\cos^{-1}(1-\frac{L^2}{2r_2^2})-\frac{1}{2}Lr_2 \sqrt{1-\frac{L^2}{4r_2^2}} $$ and thus the area of the overlap of the two circles (green region in you diagram) is $$ \frac{1}{2}[r_2^2\cos^{-1}( 1-\frac{L^2}{2r_2^2})+r_1^2\cos^{-1}(1-\frac{L^2}{2r_1^2})-L(r_2 \sqrt{1-\frac{L^2}{4r_2^2}} + r_1 \sqrt{1-\frac{L^2}{4r_1^2}})] $$ The area in the region
of your diagram is $$ \int_{x_B=-r_1}^{x_P} (\sqrt{r_2^2 -(x+x_C)^2}-y_C)dx-\int_{x_B=-r_1}^{x_P} (\sqrt{r_1^2 -x^2})dx\\ =\frac{1}{2}(x_C + x_P)\sqrt{r_2^2 - (x_C + x_P)^2}-\frac{1}{2}(x_C - r_1)\sqrt{r_2^2 - (x_C + r_1)^2}\\ -y_C (x_P + r_1)\\ - \frac{1}{2}\sqrt{r_1^2 - x_P^2} + \frac{1}{2}r_2^2\tan^{-1}(\frac{x_C + x_P}{\sqrt{r_2^2 - (x_C + x_P)^2}})\\ -\frac{1}{2}r_2^2\tan^{-1}(\frac{x_C - r_1}{\sqrt{r_2^2 - (x_C - r_1)^2}}) + \frac{1}{2}r_1^2\tan^{-1}(\frac{x_P}{\sqrt{r_1^2-x_P^2}}) + \frac{r_1^2\pi}{4} $$