Dtyjlui

please can you help to provide the mathematical steps required to calculate the area of a shape formed by two circles of different diameter joined together by two tangential lines.

Hint: look at a circle $C(O,r)$ with tangential points $T_1, T_2$. Since $OT_1$ is ortogonal to the line passing through $T_1$, it's orthogonal to the line passing through $T_2$. But the only point with this property is $T_2$. So for the circles $C, C'$, we have that $T_1T_2T_2'T_1'$ is a rectangle. So such case only occurs when the circles have the same radius. Calculate the area by area of the circle + area of the rectangle.

Also I think that by "parallel" you meant not-crossing segments. If that's the case you'll need analytic geometry or some trigonometry.

Edit: original question had by title "parallel lines". I'll suppose you have the distance of the corresponding centers and the radius of each circle.

Remark: the image has a small error regarding the final formula because I forgot that the angles the circles make are different, I divided by $pi$ for no reason and apparently $r_1^2 + r_2^2 = (r_1+r_2)^2$.

Suppose WLOG that $r_1 geq r_2$. Join the centers by a segment of given length $d$. Trace the parallels to the corresponding tangential lines from the smaller circle passing through its center. You'll get 2 rectangles of with one side being $r_2$ and the other one being the remaining side of a right-triangle with hypothenuse $d$ and side $r_1 - r_2$. Let $theta$ be the angle formed by the new parallels to the center of the smaller circumference; this angle has sine $frac{r_1 - r_2}{d}$. Find the hypothenuse with the Pythagorean theorem, calculate the area of the triangles and the rectangles and finally add the remaining circular sectors ( $frac{pi - 2theta}{2}r_2^2 + frac{2pi - 2(frac{pi}{2} - theta)}{2}{r_1}^2 $) to get the desired formula:

$$ A(r_1, r_2, d) = sqrt{d^2 - (r_1 - r_2)^2}(r_1+r_2) + sum_{cyc}(pi - 2sin^{-1}frac{r_1 - r_2}{d})r_1^2.$$

Fun fact: even with the supposition of a form of inequality, we could achieve a nice, symmetrical formula on $r_1, r_2$ as usually expected.

The two diameters are not sufficient to calculate this, as the circles can be arbitrarily far away. Let's assume we also know the distance between the centers. Denote it $L$. Denote $R_1$ and $R_2$ the radii of the circles.

Call the centers $O$ and $P$, and the points where one of the tangent line touches the two circles $A$ and $B$ respectively. Then OPAB is a trapezoid, since OA and PB are both orthogonal to AB.

Then your area can be split into a two trapezoids (OPAB and the symmetrical one) and two circle sectors.

We have $AB = sqrt{L^2 - (R_1 - R_2)^2}$ (Pythagorean theorem) and $S_{OPAB} = frac{OA + PB}{2}times AB$.

Further, we have $cos angle POA = frac{OP}{OA - PB} = frac{L}{R_1 - R_2}$.

The areas of the circle sectors are $(pi - angle POA) R_1^2$ and $(angle POA) R_2^2$.

This, overall the area is

$$(OA + PB)times sqrt{L^2 - (R_1 - R_2)^2} + (pi - arccos frac{L}{R_1 - R_2}) R_1^2 + arccos frac{L}{R_1 - R_2} R_2^2$$