What is radius of smaller circle in this case? [closed]











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There is one quarter circle have radius same as side of square. There is semi-circle that has diameter equal to side of square. Then find radius of circle which is tangent to these circle and side of square.



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closed as off-topic by José Carlos Santos, Christopher, supinf, Adrian Keister, ccorn Nov 20 at 15:59


This question appears to be off-topic. The users who voted to close gave this specific reason:


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If this question can be reworded to fit the rules in the help center, please edit the question.

















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    There is one quarter circle have radius same as side of square. There is semi-circle that has diameter equal to side of square. Then find radius of circle which is tangent to these circle and side of square.



    enter image description here










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    closed as off-topic by José Carlos Santos, Christopher, supinf, Adrian Keister, ccorn Nov 20 at 15:59


    This question appears to be off-topic. The users who voted to close gave this specific reason:


    • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, Christopher, supinf, Adrian Keister, ccorn

    If this question can be reworded to fit the rules in the help center, please edit the question.















      up vote
      -1
      down vote

      favorite









      up vote
      -1
      down vote

      favorite











      There is one quarter circle have radius same as side of square. There is semi-circle that has diameter equal to side of square. Then find radius of circle which is tangent to these circle and side of square.



      enter image description here










      share|cite|improve this question













      There is one quarter circle have radius same as side of square. There is semi-circle that has diameter equal to side of square. Then find radius of circle which is tangent to these circle and side of square.



      enter image description here







      geometry






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      asked Nov 20 at 11:24









      Angelus Mortis

      1135




      1135




      closed as off-topic by José Carlos Santos, Christopher, supinf, Adrian Keister, ccorn Nov 20 at 15:59


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, Christopher, supinf, Adrian Keister, ccorn

      If this question can be reworded to fit the rules in the help center, please edit the question.




      closed as off-topic by José Carlos Santos, Christopher, supinf, Adrian Keister, ccorn Nov 20 at 15:59


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, Christopher, supinf, Adrian Keister, ccorn

      If this question can be reworded to fit the rules in the help center, please edit the question.






















          3 Answers
          3






          active

          oldest

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          up vote
          2
          down vote



          accepted










          Hint: For each of the two large circles, construct the right triangle with hypotenuse connecting the center to the center of the small circle and legs parallel to the sides of the square.



          Let $x$ denote the distance between the lower left corner of the square and the point of tangency between the bottom side and the small circle.



          Pythagorus gives us two equations:



          $$(4-r)^2+x^2=(4+r)^2quad & quad (2-r)^2+(4-x)^2=(2+r)^2$$






          share|cite|improve this answer























          • This is an excellent hint, accessible even for student with a basic knowledge of (analytic) geometry (+1)
            – user376343
            Nov 20 at 13:37


















          up vote
          1
          down vote













          A geometric construction based on the use of conic sections:




          • Locus of the centers of circles touching two given circles is a hyperbola, whose foci are the centers of the given circles as explained here.

          • Locus of the centers of circles touching a given circle and a straight line is a parabola. Its focus is located in the center of the given circle, and the directrix is parallel to the given line, the given line being equidistant to the focus and directrix.

            In the enclosed figure is the focus $A$ and the straight line $CD.$ The directrix lies bellow, and is out of the picture.


          GeoGebra gives a nice picture, but the computation of the radius based on this method would be rather boring.



          enter image description here






          share|cite|improve this answer




























            up vote
            1
            down vote













            Hint: My attempt would be to formulate the equations for the three circles and the line segement and then check if there is a common solution possible. (Two intersection points with the circles, one with the line)



            Circle: $(x - x_0)^2 + (y - y_0)^2 = r^2$

            Line: $A x + B y = C$



            Looks close to Descartes theorem, by the way.



            Update:



            Choosing the lower left corner as origin we get the equations:



            Large circle
            $$
            x^2 + (y - 4)^2 = 4^2 quadquad (L)
            $$



            Medium sized circle
            $$
            (x-4)^2 + (y-2)^2 = 2^2 quadquad (M)
            $$



            Small circle
            $$
            (x-x_0)^2 + (y-y_0)^2 = r^2 quadquad (S)
            $$



            Line segment
            $$
            x in [0, 4] quad y = 0 quadquad (LS)
            $$



            The small circle touches the line segment from above, if the center is
            $$
            (x_0, y_0) = (x_0, r)
            $$

            for some $x_0$.






            share|cite|improve this answer























            • I thought about Descartes theorem too but it won't apply here. Anyways I quite didn't understand your hint. sorry
              – Angelus Mortis
              Nov 20 at 12:45










            • I will expand it a bit later today.
              – mvw
              Nov 20 at 15:40


















            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            2
            down vote



            accepted










            Hint: For each of the two large circles, construct the right triangle with hypotenuse connecting the center to the center of the small circle and legs parallel to the sides of the square.



            Let $x$ denote the distance between the lower left corner of the square and the point of tangency between the bottom side and the small circle.



            Pythagorus gives us two equations:



            $$(4-r)^2+x^2=(4+r)^2quad & quad (2-r)^2+(4-x)^2=(2+r)^2$$






            share|cite|improve this answer























            • This is an excellent hint, accessible even for student with a basic knowledge of (analytic) geometry (+1)
              – user376343
              Nov 20 at 13:37















            up vote
            2
            down vote



            accepted










            Hint: For each of the two large circles, construct the right triangle with hypotenuse connecting the center to the center of the small circle and legs parallel to the sides of the square.



            Let $x$ denote the distance between the lower left corner of the square and the point of tangency between the bottom side and the small circle.



            Pythagorus gives us two equations:



            $$(4-r)^2+x^2=(4+r)^2quad & quad (2-r)^2+(4-x)^2=(2+r)^2$$






            share|cite|improve this answer























            • This is an excellent hint, accessible even for student with a basic knowledge of (analytic) geometry (+1)
              – user376343
              Nov 20 at 13:37













            up vote
            2
            down vote



            accepted







            up vote
            2
            down vote



            accepted






            Hint: For each of the two large circles, construct the right triangle with hypotenuse connecting the center to the center of the small circle and legs parallel to the sides of the square.



            Let $x$ denote the distance between the lower left corner of the square and the point of tangency between the bottom side and the small circle.



            Pythagorus gives us two equations:



            $$(4-r)^2+x^2=(4+r)^2quad & quad (2-r)^2+(4-x)^2=(2+r)^2$$






            share|cite|improve this answer














            Hint: For each of the two large circles, construct the right triangle with hypotenuse connecting the center to the center of the small circle and legs parallel to the sides of the square.



            Let $x$ denote the distance between the lower left corner of the square and the point of tangency between the bottom side and the small circle.



            Pythagorus gives us two equations:



            $$(4-r)^2+x^2=(4+r)^2quad & quad (2-r)^2+(4-x)^2=(2+r)^2$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 20 at 13:50

























            answered Nov 20 at 11:31









            lulu

            38.4k24476




            38.4k24476












            • This is an excellent hint, accessible even for student with a basic knowledge of (analytic) geometry (+1)
              – user376343
              Nov 20 at 13:37


















            • This is an excellent hint, accessible even for student with a basic knowledge of (analytic) geometry (+1)
              – user376343
              Nov 20 at 13:37
















            This is an excellent hint, accessible even for student with a basic knowledge of (analytic) geometry (+1)
            – user376343
            Nov 20 at 13:37




            This is an excellent hint, accessible even for student with a basic knowledge of (analytic) geometry (+1)
            – user376343
            Nov 20 at 13:37










            up vote
            1
            down vote













            A geometric construction based on the use of conic sections:




            • Locus of the centers of circles touching two given circles is a hyperbola, whose foci are the centers of the given circles as explained here.

            • Locus of the centers of circles touching a given circle and a straight line is a parabola. Its focus is located in the center of the given circle, and the directrix is parallel to the given line, the given line being equidistant to the focus and directrix.

              In the enclosed figure is the focus $A$ and the straight line $CD.$ The directrix lies bellow, and is out of the picture.


            GeoGebra gives a nice picture, but the computation of the radius based on this method would be rather boring.



            enter image description here






            share|cite|improve this answer

























              up vote
              1
              down vote













              A geometric construction based on the use of conic sections:




              • Locus of the centers of circles touching two given circles is a hyperbola, whose foci are the centers of the given circles as explained here.

              • Locus of the centers of circles touching a given circle and a straight line is a parabola. Its focus is located in the center of the given circle, and the directrix is parallel to the given line, the given line being equidistant to the focus and directrix.

                In the enclosed figure is the focus $A$ and the straight line $CD.$ The directrix lies bellow, and is out of the picture.


              GeoGebra gives a nice picture, but the computation of the radius based on this method would be rather boring.



              enter image description here






              share|cite|improve this answer























                up vote
                1
                down vote










                up vote
                1
                down vote









                A geometric construction based on the use of conic sections:




                • Locus of the centers of circles touching two given circles is a hyperbola, whose foci are the centers of the given circles as explained here.

                • Locus of the centers of circles touching a given circle and a straight line is a parabola. Its focus is located in the center of the given circle, and the directrix is parallel to the given line, the given line being equidistant to the focus and directrix.

                  In the enclosed figure is the focus $A$ and the straight line $CD.$ The directrix lies bellow, and is out of the picture.


                GeoGebra gives a nice picture, but the computation of the radius based on this method would be rather boring.



                enter image description here






                share|cite|improve this answer












                A geometric construction based on the use of conic sections:




                • Locus of the centers of circles touching two given circles is a hyperbola, whose foci are the centers of the given circles as explained here.

                • Locus of the centers of circles touching a given circle and a straight line is a parabola. Its focus is located in the center of the given circle, and the directrix is parallel to the given line, the given line being equidistant to the focus and directrix.

                  In the enclosed figure is the focus $A$ and the straight line $CD.$ The directrix lies bellow, and is out of the picture.


                GeoGebra gives a nice picture, but the computation of the radius based on this method would be rather boring.



                enter image description here







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 20 at 14:00









                user376343

                2,6332818




                2,6332818






















                    up vote
                    1
                    down vote













                    Hint: My attempt would be to formulate the equations for the three circles and the line segement and then check if there is a common solution possible. (Two intersection points with the circles, one with the line)



                    Circle: $(x - x_0)^2 + (y - y_0)^2 = r^2$

                    Line: $A x + B y = C$



                    Looks close to Descartes theorem, by the way.



                    Update:



                    Choosing the lower left corner as origin we get the equations:



                    Large circle
                    $$
                    x^2 + (y - 4)^2 = 4^2 quadquad (L)
                    $$



                    Medium sized circle
                    $$
                    (x-4)^2 + (y-2)^2 = 2^2 quadquad (M)
                    $$



                    Small circle
                    $$
                    (x-x_0)^2 + (y-y_0)^2 = r^2 quadquad (S)
                    $$



                    Line segment
                    $$
                    x in [0, 4] quad y = 0 quadquad (LS)
                    $$



                    The small circle touches the line segment from above, if the center is
                    $$
                    (x_0, y_0) = (x_0, r)
                    $$

                    for some $x_0$.






                    share|cite|improve this answer























                    • I thought about Descartes theorem too but it won't apply here. Anyways I quite didn't understand your hint. sorry
                      – Angelus Mortis
                      Nov 20 at 12:45










                    • I will expand it a bit later today.
                      – mvw
                      Nov 20 at 15:40















                    up vote
                    1
                    down vote













                    Hint: My attempt would be to formulate the equations for the three circles and the line segement and then check if there is a common solution possible. (Two intersection points with the circles, one with the line)



                    Circle: $(x - x_0)^2 + (y - y_0)^2 = r^2$

                    Line: $A x + B y = C$



                    Looks close to Descartes theorem, by the way.



                    Update:



                    Choosing the lower left corner as origin we get the equations:



                    Large circle
                    $$
                    x^2 + (y - 4)^2 = 4^2 quadquad (L)
                    $$



                    Medium sized circle
                    $$
                    (x-4)^2 + (y-2)^2 = 2^2 quadquad (M)
                    $$



                    Small circle
                    $$
                    (x-x_0)^2 + (y-y_0)^2 = r^2 quadquad (S)
                    $$



                    Line segment
                    $$
                    x in [0, 4] quad y = 0 quadquad (LS)
                    $$



                    The small circle touches the line segment from above, if the center is
                    $$
                    (x_0, y_0) = (x_0, r)
                    $$

                    for some $x_0$.






                    share|cite|improve this answer























                    • I thought about Descartes theorem too but it won't apply here. Anyways I quite didn't understand your hint. sorry
                      – Angelus Mortis
                      Nov 20 at 12:45










                    • I will expand it a bit later today.
                      – mvw
                      Nov 20 at 15:40













                    up vote
                    1
                    down vote










                    up vote
                    1
                    down vote









                    Hint: My attempt would be to formulate the equations for the three circles and the line segement and then check if there is a common solution possible. (Two intersection points with the circles, one with the line)



                    Circle: $(x - x_0)^2 + (y - y_0)^2 = r^2$

                    Line: $A x + B y = C$



                    Looks close to Descartes theorem, by the way.



                    Update:



                    Choosing the lower left corner as origin we get the equations:



                    Large circle
                    $$
                    x^2 + (y - 4)^2 = 4^2 quadquad (L)
                    $$



                    Medium sized circle
                    $$
                    (x-4)^2 + (y-2)^2 = 2^2 quadquad (M)
                    $$



                    Small circle
                    $$
                    (x-x_0)^2 + (y-y_0)^2 = r^2 quadquad (S)
                    $$



                    Line segment
                    $$
                    x in [0, 4] quad y = 0 quadquad (LS)
                    $$



                    The small circle touches the line segment from above, if the center is
                    $$
                    (x_0, y_0) = (x_0, r)
                    $$

                    for some $x_0$.






                    share|cite|improve this answer














                    Hint: My attempt would be to formulate the equations for the three circles and the line segement and then check if there is a common solution possible. (Two intersection points with the circles, one with the line)



                    Circle: $(x - x_0)^2 + (y - y_0)^2 = r^2$

                    Line: $A x + B y = C$



                    Looks close to Descartes theorem, by the way.



                    Update:



                    Choosing the lower left corner as origin we get the equations:



                    Large circle
                    $$
                    x^2 + (y - 4)^2 = 4^2 quadquad (L)
                    $$



                    Medium sized circle
                    $$
                    (x-4)^2 + (y-2)^2 = 2^2 quadquad (M)
                    $$



                    Small circle
                    $$
                    (x-x_0)^2 + (y-y_0)^2 = r^2 quadquad (S)
                    $$



                    Line segment
                    $$
                    x in [0, 4] quad y = 0 quadquad (LS)
                    $$



                    The small circle touches the line segment from above, if the center is
                    $$
                    (x_0, y_0) = (x_0, r)
                    $$

                    for some $x_0$.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Nov 20 at 15:39

























                    answered Nov 20 at 11:32









                    mvw

                    31.2k22252




                    31.2k22252












                    • I thought about Descartes theorem too but it won't apply here. Anyways I quite didn't understand your hint. sorry
                      – Angelus Mortis
                      Nov 20 at 12:45










                    • I will expand it a bit later today.
                      – mvw
                      Nov 20 at 15:40


















                    • I thought about Descartes theorem too but it won't apply here. Anyways I quite didn't understand your hint. sorry
                      – Angelus Mortis
                      Nov 20 at 12:45










                    • I will expand it a bit later today.
                      – mvw
                      Nov 20 at 15:40
















                    I thought about Descartes theorem too but it won't apply here. Anyways I quite didn't understand your hint. sorry
                    – Angelus Mortis
                    Nov 20 at 12:45




                    I thought about Descartes theorem too but it won't apply here. Anyways I quite didn't understand your hint. sorry
                    – Angelus Mortis
                    Nov 20 at 12:45












                    I will expand it a bit later today.
                    – mvw
                    Nov 20 at 15:40




                    I will expand it a bit later today.
                    – mvw
                    Nov 20 at 15:40



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