Mixing
How do you calculate area of two circles joined by tangent lines
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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.
area
edited Jan 4 at 14:47
DonielF
484515
asked Jan 4 at 12:21
Karl WhopplesKarl Whopples
1
$endgroup$
add a comment |
$begingroup$
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.
area
edited Jan 4 at 14:47
DonielF
484515
asked Jan 4 at 12:21
Karl WhopplesKarl Whopples
1
$endgroup$
3
$begingroup$
What are the givens? Do you know the diameters of the circles, or the lengths of the tangents, or anything like that?
$endgroup$
– MJD
Jan 4 at 12:30
$begingroup$
More information for you: The larger circle has a diameter 36mm the smaller circle a diameter of 28mm. The distance between the two circle centers is 18.5mm.
$endgroup$
– Karl Whopples
Jan 7 at 9:34
add a comment |
$begingroup$
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.
area
edited Jan 4 at 14:47
DonielF
484515
asked Jan 4 at 12:21
Karl WhopplesKarl Whopples
1
$endgroup$
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.
area
area
edited Jan 4 at 14:47
DonielF
484515
asked Jan 4 at 12:21
Karl WhopplesKarl Whopples
1
edited Jan 4 at 14:47
DonielF
484515
asked Jan 4 at 12:21
Karl WhopplesKarl Whopples
1
edited Jan 4 at 14:47
DonielF
484515
edited Jan 4 at 14:47
DonielF
484515
edited Jan 4 at 14:47
DonielF
484515
484515
asked Jan 4 at 12:21
Karl WhopplesKarl Whopples
1
asked Jan 4 at 12:21
Karl WhopplesKarl Whopples
1
asked Jan 4 at 12:21
Karl WhopplesKarl Whopples
1
1
3
$begingroup$
What are the givens? Do you know the diameters of the circles, or the lengths of the tangents, or anything like that?
$endgroup$
– MJD
Jan 4 at 12:30
$begingroup$
More information for you: The larger circle has a diameter 36mm the smaller circle a diameter of 28mm. The distance between the two circle centers is 18.5mm.
$endgroup$
– Karl Whopples
Jan 7 at 9:34
add a comment |
3
$begingroup$
What are the givens? Do you know the diameters of the circles, or the lengths of the tangents, or anything like that?
$endgroup$
– MJD
Jan 4 at 12:30
$begingroup$
More information for you: The larger circle has a diameter 36mm the smaller circle a diameter of 28mm. The distance between the two circle centers is 18.5mm.
$endgroup$
– Karl Whopples
Jan 7 at 9:34
3
3
$begingroup$
What are the givens? Do you know the diameters of the circles, or the lengths of the tangents, or anything like that?
$endgroup$
– MJD
Jan 4 at 12:30
$begingroup$
What are the givens? Do you know the diameters of the circles, or the lengths of the tangents, or anything like that?
$endgroup$
– MJD
Jan 4 at 12:30
$begingroup$
More information for you: The larger circle has a diameter 36mm the smaller circle a diameter of 28mm. The distance between the two circle centers is 18.5mm.
$endgroup$
– Karl Whopples
Jan 7 at 9:34
$begingroup$
More information for you: The larger circle has a diameter 36mm the smaller circle a diameter of 28mm. The distance between the two circle centers is 18.5mm.
$endgroup$
– Karl Whopples
Jan 7 at 9:34
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
answered Jan 4 at 12:32
Lucas HenriqueLucas Henrique
1,024414
$endgroup$
add a comment |
$begingroup$
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
$$(OA + PB)times sqrt{L^2 - (R_1 - R_2)^2} + (pi - arccos frac{L}{R_1 - R_2}) R_1^2 + arccos frac{L}{R_1 - R_2} R_2^2$$
answered Jan 4 at 12:44
Todor MarkovTodor Markov
1,819410
$endgroup$
$begingroup$
The question is not clear about its symbol definitions, but I think $D_1$ and $D_2$ were intended to be the diameters of the circles, not the radii.
$endgroup$
– David K
Jan 4 at 14:17
$begingroup$
Hi, thanks for taking the time to try to resolve this problem. I can now provide some measurements which in fact will result in the circles partially overlapping. The diameter of the larger circle is 36mm and the diameter of the smaller circle is 28mm. The distance between the circle centers is 18.5mm.
$endgroup$
– Karl Whopples
Jan 7 at 10:54
$begingroup$
@KarlWhopples That shouldn't affect the calculation. The circle sectors I'm adding up shouldn't overlap even if the circles themselves do.
$endgroup$
– Todor Markov
Jan 7 at 12:16
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
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$begingroup$
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.
answered Jan 4 at 12:32
Lucas HenriqueLucas Henrique
1,024414
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 4 at 12:32
Lucas HenriqueLucas Henrique
1,024414
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 4 at 12:32
Lucas HenriqueLucas Henrique
1,024414
$endgroup$
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.
answered Jan 4 at 12:32
Lucas HenriqueLucas Henrique
1,024414
edited Jan 4 at 13:25
answered Jan 4 at 12:32
Lucas HenriqueLucas Henrique
1,024414
answered Jan 4 at 12:32
Lucas HenriqueLucas Henrique
1,024414
answered Jan 4 at 12:32
Lucas HenriqueLucas Henrique
1,024414
1,024414
add a comment |
add a comment |
$begingroup$
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
$$(OA + PB)times sqrt{L^2 - (R_1 - R_2)^2} + (pi - arccos frac{L}{R_1 - R_2}) R_1^2 + arccos frac{L}{R_1 - R_2} R_2^2$$
answered Jan 4 at 12:44
Todor MarkovTodor Markov
1,819410
$endgroup$
$begingroup$
The question is not clear about its symbol definitions, but I think $D_1$ and $D_2$ were intended to be the diameters of the circles, not the radii.
$endgroup$
– David K
Jan 4 at 14:17
$begingroup$
Hi, thanks for taking the time to try to resolve this problem. I can now provide some measurements which in fact will result in the circles partially overlapping. The diameter of the larger circle is 36mm and the diameter of the smaller circle is 28mm. The distance between the circle centers is 18.5mm.
$endgroup$
– Karl Whopples
Jan 7 at 10:54
$begingroup$
@KarlWhopples That shouldn't affect the calculation. The circle sectors I'm adding up shouldn't overlap even if the circles themselves do.
$endgroup$
– Todor Markov
Jan 7 at 12:16
add a comment |
$begingroup$
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
$$(OA + PB)times sqrt{L^2 - (R_1 - R_2)^2} + (pi - arccos frac{L}{R_1 - R_2}) R_1^2 + arccos frac{L}{R_1 - R_2} R_2^2$$
answered Jan 4 at 12:44
Todor MarkovTodor Markov
1,819410
$endgroup$
$begingroup$
The question is not clear about its symbol definitions, but I think $D_1$ and $D_2$ were intended to be the diameters of the circles, not the radii.
$endgroup$
– David K
Jan 4 at 14:17
$begingroup$
Hi, thanks for taking the time to try to resolve this problem. I can now provide some measurements which in fact will result in the circles partially overlapping. The diameter of the larger circle is 36mm and the diameter of the smaller circle is 28mm. The distance between the circle centers is 18.5mm.
$endgroup$
– Karl Whopples
Jan 7 at 10:54
$begingroup$
@KarlWhopples That shouldn't affect the calculation. The circle sectors I'm adding up shouldn't overlap even if the circles themselves do.
$endgroup$
– Todor Markov
Jan 7 at 12:16
add a comment |
$begingroup$
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
$$(OA + PB)times sqrt{L^2 - (R_1 - R_2)^2} + (pi - arccos frac{L}{R_1 - R_2}) R_1^2 + arccos frac{L}{R_1 - R_2} R_2^2$$
answered Jan 4 at 12:44
Todor MarkovTodor Markov
1,819410
$endgroup$
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
$$(OA + PB)times sqrt{L^2 - (R_1 - R_2)^2} + (pi - arccos frac{L}{R_1 - R_2}) R_1^2 + arccos frac{L}{R_1 - R_2} R_2^2$$
answered Jan 4 at 12:44
Todor MarkovTodor Markov
1,819410
edited Jan 4 at 14:39
answered Jan 4 at 12:44
Todor MarkovTodor Markov
1,819410
answered Jan 4 at 12:44
Todor MarkovTodor Markov
1,819410
answered Jan 4 at 12:44
Todor MarkovTodor Markov
1,819410
1,819410
$begingroup$
The question is not clear about its symbol definitions, but I think $D_1$ and $D_2$ were intended to be the diameters of the circles, not the radii.
$endgroup$
– David K
Jan 4 at 14:17
$begingroup$
Hi, thanks for taking the time to try to resolve this problem. I can now provide some measurements which in fact will result in the circles partially overlapping. The diameter of the larger circle is 36mm and the diameter of the smaller circle is 28mm. The distance between the circle centers is 18.5mm.
$endgroup$
– Karl Whopples
Jan 7 at 10:54
$begingroup$
@KarlWhopples That shouldn't affect the calculation. The circle sectors I'm adding up shouldn't overlap even if the circles themselves do.
$endgroup$
– Todor Markov
Jan 7 at 12:16
add a comment |
$begingroup$
The question is not clear about its symbol definitions, but I think $D_1$ and $D_2$ were intended to be the diameters of the circles, not the radii.
$endgroup$
– David K
Jan 4 at 14:17
$begingroup$
Hi, thanks for taking the time to try to resolve this problem. I can now provide some measurements which in fact will result in the circles partially overlapping. The diameter of the larger circle is 36mm and the diameter of the smaller circle is 28mm. The distance between the circle centers is 18.5mm.
$endgroup$
– Karl Whopples
Jan 7 at 10:54
$begingroup$
@KarlWhopples That shouldn't affect the calculation. The circle sectors I'm adding up shouldn't overlap even if the circles themselves do.
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– Todor Markov
Jan 7 at 12:16
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The question is not clear about its symbol definitions, but I think $D_1$ and $D_2$ were intended to be the diameters of the circles, not the radii.
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– David K
Jan 4 at 14:17
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The question is not clear about its symbol definitions, but I think $D_1$ and $D_2$ were intended to be the diameters of the circles, not the radii.
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– David K
Jan 4 at 14:17
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Hi, thanks for taking the time to try to resolve this problem. I can now provide some measurements which in fact will result in the circles partially overlapping. The diameter of the larger circle is 36mm and the diameter of the smaller circle is 28mm. The distance between the circle centers is 18.5mm.
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– Karl Whopples
Jan 7 at 10:54
$begingroup$
Hi, thanks for taking the time to try to resolve this problem. I can now provide some measurements which in fact will result in the circles partially overlapping. The diameter of the larger circle is 36mm and the diameter of the smaller circle is 28mm. The distance between the circle centers is 18.5mm.
$endgroup$
– Karl Whopples
Jan 7 at 10:54
$begingroup$
@KarlWhopples That shouldn't affect the calculation. The circle sectors I'm adding up shouldn't overlap even if the circles themselves do.
$endgroup$
– Todor Markov
Jan 7 at 12:16
$begingroup$
@KarlWhopples That shouldn't affect the calculation. The circle sectors I'm adding up shouldn't overlap even if the circles themselves do.
$endgroup$
– Todor Markov
Jan 7 at 12:16
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$begingroup$
What are the givens? Do you know the diameters of the circles, or the lengths of the tangents, or anything like that?
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– MJD
Jan 4 at 12:30
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More information for you: The larger circle has a diameter 36mm the smaller circle a diameter of 28mm. The distance between the two circle centers is 18.5mm.
$endgroup$
– Karl Whopples
Jan 7 at 9:34