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.



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    What are the givens? Do you know the diameters of the circles, or the lengths of the tangents, or anything like that?
    – MJD
    Jan 4 at 12:30










  • 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.
    – Karl Whopples
    yesterday
















0














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.



Shape










share|cite|improve this question









New contributor




Karl Whopples is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 3




    What are the givens? Do you know the diameters of the circles, or the lengths of the tangents, or anything like that?
    – MJD
    Jan 4 at 12:30










  • 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.
    – Karl Whopples
    yesterday














0












0








0







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.



Shape










share|cite|improve this question









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Karl Whopples is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











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.



Shape







area






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Karl Whopples is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited Jan 4 at 14:47









DonielF

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asked Jan 4 at 12:21









Karl WhopplesKarl Whopples

1




1




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Karl Whopples is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Karl Whopples is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 3




    What are the givens? Do you know the diameters of the circles, or the lengths of the tangents, or anything like that?
    – MJD
    Jan 4 at 12:30










  • 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.
    – Karl Whopples
    yesterday














  • 3




    What are the givens? Do you know the diameters of the circles, or the lengths of the tangents, or anything like that?
    – MJD
    Jan 4 at 12:30










  • 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.
    – Karl Whopples
    yesterday








3




3




What are the givens? Do you know the diameters of the circles, or the lengths of the tangents, or anything like that?
– MJD
Jan 4 at 12:30




What are the givens? Do you know the diameters of the circles, or the lengths of the tangents, or anything like that?
– MJD
Jan 4 at 12:30












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.
– Karl Whopples
yesterday




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.
– Karl Whopples
yesterday










2 Answers
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0














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 $r_1^2 + r_2^2 = (r_1+r_2)^2$.
Geometric constructions



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.






share|cite|improve this answer































    0














    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



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






    share|cite|improve this answer























    • 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.
      – David K
      Jan 4 at 14:17










    • 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.
      – Karl Whopples
      yesterday










    • @KarlWhopples That shouldn't affect the calculation. The circle sectors I'm adding up shouldn't overlap even if the circles themselves do.
      – Todor Markov
      yesterday











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    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 $r_1^2 + r_2^2 = (r_1+r_2)^2$.
    Geometric constructions



    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.






    share|cite|improve this answer




























      0














      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 $r_1^2 + r_2^2 = (r_1+r_2)^2$.
      Geometric constructions



      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.






      share|cite|improve this answer


























        0












        0








        0






        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 $r_1^2 + r_2^2 = (r_1+r_2)^2$.
        Geometric constructions



        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.






        share|cite|improve this answer














        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 $r_1^2 + r_2^2 = (r_1+r_2)^2$.
        Geometric constructions



        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.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 4 at 13:25

























        answered Jan 4 at 12:32









        Lucas HenriqueLucas Henrique

        968414




        968414























            0














            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



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






            share|cite|improve this answer























            • 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.
              – David K
              Jan 4 at 14:17










            • 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.
              – Karl Whopples
              yesterday










            • @KarlWhopples That shouldn't affect the calculation. The circle sectors I'm adding up shouldn't overlap even if the circles themselves do.
              – Todor Markov
              yesterday
















            0














            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



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






            share|cite|improve this answer























            • 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.
              – David K
              Jan 4 at 14:17










            • 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.
              – Karl Whopples
              yesterday










            • @KarlWhopples That shouldn't affect the calculation. The circle sectors I'm adding up shouldn't overlap even if the circles themselves do.
              – Todor Markov
              yesterday














            0












            0








            0






            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



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






            share|cite|improve this answer














            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



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







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 4 at 14:39

























            answered Jan 4 at 12:44









            Todor MarkovTodor Markov

            1,664410




            1,664410












            • 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.
              – David K
              Jan 4 at 14:17










            • 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.
              – Karl Whopples
              yesterday










            • @KarlWhopples That shouldn't affect the calculation. The circle sectors I'm adding up shouldn't overlap even if the circles themselves do.
              – Todor Markov
              yesterday


















            • 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.
              – David K
              Jan 4 at 14:17










            • 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.
              – Karl Whopples
              yesterday










            • @KarlWhopples That shouldn't affect the calculation. The circle sectors I'm adding up shouldn't overlap even if the circles themselves do.
              – Todor Markov
              yesterday
















            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.
            – David K
            Jan 4 at 14:17




            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.
            – David K
            Jan 4 at 14:17












            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.
            – Karl Whopples
            yesterday




            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.
            – Karl Whopples
            yesterday












            @KarlWhopples That shouldn't affect the calculation. The circle sectors I'm adding up shouldn't overlap even if the circles themselves do.
            – Todor Markov
            yesterday




            @KarlWhopples That shouldn't affect the calculation. The circle sectors I'm adding up shouldn't overlap even if the circles themselves do.
            – Todor Markov
            yesterday










            Karl Whopples is a new contributor. Be nice, and check out our Code of Conduct.










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            Karl Whopples is a new contributor. Be nice, and check out our Code of Conduct.













            Karl Whopples is a new contributor. Be nice, and check out our Code of Conduct.












            Karl Whopples is a new contributor. Be nice, and check out our Code of Conduct.
















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