How do you calculate area of two circles joined by tangent lines

Question

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.

Answer 1

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 - 2\theta}{2}r_2^2 + \frac{2\pi - 2(\frac{\pi}{2} - \theta)}{2}{r_1}^2 $) to get the desired formula:

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

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.

Answer 2

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

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