Circle circumference point calculation












0












$begingroup$


In the image below, I have a part of a circle. Given,



$$text{chord }d=1050 mm\
text{height }f=50 mm\
text{radius }R=2781 mm\
text{centre }O(700 mm,2781 mm)$$



and $3$ points $A(0,200), B(1050,200), C(525,150)$.




I would like to know if there is an equation to calculate the $y$ coordinate for any given $x$ coordinate, for example the point $D(200,Y)$.




https://i.stack.imgur.com/VzNUf.png










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  • $begingroup$
    I would start by figuring out the equation of the circle. I mean, in the form $$ (x-x_0)^2 + (y-y_0)^2 = R^2 $$
    $endgroup$
    – Matti P.
    Jan 14 at 13:20










  • $begingroup$
    The question has already been asked here.
    $endgroup$
    – whiskeyo
    Jan 14 at 13:22


















0












$begingroup$


In the image below, I have a part of a circle. Given,



$$text{chord }d=1050 mm\
text{height }f=50 mm\
text{radius }R=2781 mm\
text{centre }O(700 mm,2781 mm)$$



and $3$ points $A(0,200), B(1050,200), C(525,150)$.




I would like to know if there is an equation to calculate the $y$ coordinate for any given $x$ coordinate, for example the point $D(200,Y)$.




https://i.stack.imgur.com/VzNUf.png










share|cite|improve this question











$endgroup$












  • $begingroup$
    I would start by figuring out the equation of the circle. I mean, in the form $$ (x-x_0)^2 + (y-y_0)^2 = R^2 $$
    $endgroup$
    – Matti P.
    Jan 14 at 13:20










  • $begingroup$
    The question has already been asked here.
    $endgroup$
    – whiskeyo
    Jan 14 at 13:22
















0












0








0





$begingroup$


In the image below, I have a part of a circle. Given,



$$text{chord }d=1050 mm\
text{height }f=50 mm\
text{radius }R=2781 mm\
text{centre }O(700 mm,2781 mm)$$



and $3$ points $A(0,200), B(1050,200), C(525,150)$.




I would like to know if there is an equation to calculate the $y$ coordinate for any given $x$ coordinate, for example the point $D(200,Y)$.




https://i.stack.imgur.com/VzNUf.png










share|cite|improve this question











$endgroup$




In the image below, I have a part of a circle. Given,



$$text{chord }d=1050 mm\
text{height }f=50 mm\
text{radius }R=2781 mm\
text{centre }O(700 mm,2781 mm)$$



and $3$ points $A(0,200), B(1050,200), C(525,150)$.




I would like to know if there is an equation to calculate the $y$ coordinate for any given $x$ coordinate, for example the point $D(200,Y)$.




https://i.stack.imgur.com/VzNUf.png







trigonometry circle






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edited Jan 14 at 13:32









Shubham Johri

5,192717




5,192717










asked Jan 14 at 13:11









AleksanderAleksander

11




11












  • $begingroup$
    I would start by figuring out the equation of the circle. I mean, in the form $$ (x-x_0)^2 + (y-y_0)^2 = R^2 $$
    $endgroup$
    – Matti P.
    Jan 14 at 13:20










  • $begingroup$
    The question has already been asked here.
    $endgroup$
    – whiskeyo
    Jan 14 at 13:22




















  • $begingroup$
    I would start by figuring out the equation of the circle. I mean, in the form $$ (x-x_0)^2 + (y-y_0)^2 = R^2 $$
    $endgroup$
    – Matti P.
    Jan 14 at 13:20










  • $begingroup$
    The question has already been asked here.
    $endgroup$
    – whiskeyo
    Jan 14 at 13:22


















$begingroup$
I would start by figuring out the equation of the circle. I mean, in the form $$ (x-x_0)^2 + (y-y_0)^2 = R^2 $$
$endgroup$
– Matti P.
Jan 14 at 13:20




$begingroup$
I would start by figuring out the equation of the circle. I mean, in the form $$ (x-x_0)^2 + (y-y_0)^2 = R^2 $$
$endgroup$
– Matti P.
Jan 14 at 13:20












$begingroup$
The question has already been asked here.
$endgroup$
– whiskeyo
Jan 14 at 13:22






$begingroup$
The question has already been asked here.
$endgroup$
– whiskeyo
Jan 14 at 13:22












1 Answer
1






active

oldest

votes


















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$begingroup$

Try to solve something more general: Suppose that you have 3 arbitrary points $A(x_A,y_A)$, $B(x_B,y_B)$ and $C(x_C,y_C)$. Let's find center $O(x_O,y_O)$ and radius $R$ of the circle passing through all three points.



Denote midpoints of segments $AB$,$BC$ with $P$,$Q$:



$$P(x_P=frac{x_A+x_B}{2}, y_P=frac{y_A+x_B}{2})$$



$$Q(x_Q=frac{x_B+x_C}{2}, y_Q=frac{y_B+y_C}{2})$$



Obviously $OPbot AB$ and $OQbot BC$ which means that:



$$(x_O-x_P)(x_B-x_A)+(y_O-y_P)(y_B-y_A)=0$$
$$(x_O-x_Q)(x_C-x_B)+(y_O-y_Q)(y_C-y_B)=0$$



or:



$$(x_O-frac{x_A+x_B}{2})(x_B-x_A)+(y_O-frac{y_A+y_B}{2})(y_B-y_A)=0$$
$$(x_O-frac{x_B+x_C}{2})(x_C-x_B)+(y_O-frac{y_B+y_C}{2})(y_C-y_B)=0$$



or:



$$(x_B-x_A)x_O+(y_B-y_A)y_O=frac{x_B^2-x_A^2+y_B^2-y_A^2}{2}tag{1}$$



$$(x_C-x_B)x_O+(y_C-y_B)y_O=frac{x_C^2-x_B^2+y_C^2-y_B^2}{2}tag{2}$$



The point is: equations (1) and (2) are linear wtih two unknowns $(x_O,y_O)$ and, assuming that points $A,B,C$ are not collinear, you can always find a unique solution. You can also come up with a general formula for $(x_O,y_O)$ but I leave it up to you as an exercise.



Once you have $x_O,y_O$ you can find the radius of the circle:



$$R=sqrt{(x_A-x_O)^2+(y_A-y_O)^2}$$



For any point on the circle $K(x_K,y_K)$:



$$(x_K-x_O)^2+(y_K-y_O)^2=R^2$$



And if you know $x_K$, you can calculate the value of $y_K$:



$$y_K=y_Opmsqrt{R^2-(x_K-x_O)^2}$$






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    1 Answer
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    0












    $begingroup$

    Try to solve something more general: Suppose that you have 3 arbitrary points $A(x_A,y_A)$, $B(x_B,y_B)$ and $C(x_C,y_C)$. Let's find center $O(x_O,y_O)$ and radius $R$ of the circle passing through all three points.



    Denote midpoints of segments $AB$,$BC$ with $P$,$Q$:



    $$P(x_P=frac{x_A+x_B}{2}, y_P=frac{y_A+x_B}{2})$$



    $$Q(x_Q=frac{x_B+x_C}{2}, y_Q=frac{y_B+y_C}{2})$$



    Obviously $OPbot AB$ and $OQbot BC$ which means that:



    $$(x_O-x_P)(x_B-x_A)+(y_O-y_P)(y_B-y_A)=0$$
    $$(x_O-x_Q)(x_C-x_B)+(y_O-y_Q)(y_C-y_B)=0$$



    or:



    $$(x_O-frac{x_A+x_B}{2})(x_B-x_A)+(y_O-frac{y_A+y_B}{2})(y_B-y_A)=0$$
    $$(x_O-frac{x_B+x_C}{2})(x_C-x_B)+(y_O-frac{y_B+y_C}{2})(y_C-y_B)=0$$



    or:



    $$(x_B-x_A)x_O+(y_B-y_A)y_O=frac{x_B^2-x_A^2+y_B^2-y_A^2}{2}tag{1}$$



    $$(x_C-x_B)x_O+(y_C-y_B)y_O=frac{x_C^2-x_B^2+y_C^2-y_B^2}{2}tag{2}$$



    The point is: equations (1) and (2) are linear wtih two unknowns $(x_O,y_O)$ and, assuming that points $A,B,C$ are not collinear, you can always find a unique solution. You can also come up with a general formula for $(x_O,y_O)$ but I leave it up to you as an exercise.



    Once you have $x_O,y_O$ you can find the radius of the circle:



    $$R=sqrt{(x_A-x_O)^2+(y_A-y_O)^2}$$



    For any point on the circle $K(x_K,y_K)$:



    $$(x_K-x_O)^2+(y_K-y_O)^2=R^2$$



    And if you know $x_K$, you can calculate the value of $y_K$:



    $$y_K=y_Opmsqrt{R^2-(x_K-x_O)^2}$$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Try to solve something more general: Suppose that you have 3 arbitrary points $A(x_A,y_A)$, $B(x_B,y_B)$ and $C(x_C,y_C)$. Let's find center $O(x_O,y_O)$ and radius $R$ of the circle passing through all three points.



      Denote midpoints of segments $AB$,$BC$ with $P$,$Q$:



      $$P(x_P=frac{x_A+x_B}{2}, y_P=frac{y_A+x_B}{2})$$



      $$Q(x_Q=frac{x_B+x_C}{2}, y_Q=frac{y_B+y_C}{2})$$



      Obviously $OPbot AB$ and $OQbot BC$ which means that:



      $$(x_O-x_P)(x_B-x_A)+(y_O-y_P)(y_B-y_A)=0$$
      $$(x_O-x_Q)(x_C-x_B)+(y_O-y_Q)(y_C-y_B)=0$$



      or:



      $$(x_O-frac{x_A+x_B}{2})(x_B-x_A)+(y_O-frac{y_A+y_B}{2})(y_B-y_A)=0$$
      $$(x_O-frac{x_B+x_C}{2})(x_C-x_B)+(y_O-frac{y_B+y_C}{2})(y_C-y_B)=0$$



      or:



      $$(x_B-x_A)x_O+(y_B-y_A)y_O=frac{x_B^2-x_A^2+y_B^2-y_A^2}{2}tag{1}$$



      $$(x_C-x_B)x_O+(y_C-y_B)y_O=frac{x_C^2-x_B^2+y_C^2-y_B^2}{2}tag{2}$$



      The point is: equations (1) and (2) are linear wtih two unknowns $(x_O,y_O)$ and, assuming that points $A,B,C$ are not collinear, you can always find a unique solution. You can also come up with a general formula for $(x_O,y_O)$ but I leave it up to you as an exercise.



      Once you have $x_O,y_O$ you can find the radius of the circle:



      $$R=sqrt{(x_A-x_O)^2+(y_A-y_O)^2}$$



      For any point on the circle $K(x_K,y_K)$:



      $$(x_K-x_O)^2+(y_K-y_O)^2=R^2$$



      And if you know $x_K$, you can calculate the value of $y_K$:



      $$y_K=y_Opmsqrt{R^2-(x_K-x_O)^2}$$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Try to solve something more general: Suppose that you have 3 arbitrary points $A(x_A,y_A)$, $B(x_B,y_B)$ and $C(x_C,y_C)$. Let's find center $O(x_O,y_O)$ and radius $R$ of the circle passing through all three points.



        Denote midpoints of segments $AB$,$BC$ with $P$,$Q$:



        $$P(x_P=frac{x_A+x_B}{2}, y_P=frac{y_A+x_B}{2})$$



        $$Q(x_Q=frac{x_B+x_C}{2}, y_Q=frac{y_B+y_C}{2})$$



        Obviously $OPbot AB$ and $OQbot BC$ which means that:



        $$(x_O-x_P)(x_B-x_A)+(y_O-y_P)(y_B-y_A)=0$$
        $$(x_O-x_Q)(x_C-x_B)+(y_O-y_Q)(y_C-y_B)=0$$



        or:



        $$(x_O-frac{x_A+x_B}{2})(x_B-x_A)+(y_O-frac{y_A+y_B}{2})(y_B-y_A)=0$$
        $$(x_O-frac{x_B+x_C}{2})(x_C-x_B)+(y_O-frac{y_B+y_C}{2})(y_C-y_B)=0$$



        or:



        $$(x_B-x_A)x_O+(y_B-y_A)y_O=frac{x_B^2-x_A^2+y_B^2-y_A^2}{2}tag{1}$$



        $$(x_C-x_B)x_O+(y_C-y_B)y_O=frac{x_C^2-x_B^2+y_C^2-y_B^2}{2}tag{2}$$



        The point is: equations (1) and (2) are linear wtih two unknowns $(x_O,y_O)$ and, assuming that points $A,B,C$ are not collinear, you can always find a unique solution. You can also come up with a general formula for $(x_O,y_O)$ but I leave it up to you as an exercise.



        Once you have $x_O,y_O$ you can find the radius of the circle:



        $$R=sqrt{(x_A-x_O)^2+(y_A-y_O)^2}$$



        For any point on the circle $K(x_K,y_K)$:



        $$(x_K-x_O)^2+(y_K-y_O)^2=R^2$$



        And if you know $x_K$, you can calculate the value of $y_K$:



        $$y_K=y_Opmsqrt{R^2-(x_K-x_O)^2}$$






        share|cite|improve this answer









        $endgroup$



        Try to solve something more general: Suppose that you have 3 arbitrary points $A(x_A,y_A)$, $B(x_B,y_B)$ and $C(x_C,y_C)$. Let's find center $O(x_O,y_O)$ and radius $R$ of the circle passing through all three points.



        Denote midpoints of segments $AB$,$BC$ with $P$,$Q$:



        $$P(x_P=frac{x_A+x_B}{2}, y_P=frac{y_A+x_B}{2})$$



        $$Q(x_Q=frac{x_B+x_C}{2}, y_Q=frac{y_B+y_C}{2})$$



        Obviously $OPbot AB$ and $OQbot BC$ which means that:



        $$(x_O-x_P)(x_B-x_A)+(y_O-y_P)(y_B-y_A)=0$$
        $$(x_O-x_Q)(x_C-x_B)+(y_O-y_Q)(y_C-y_B)=0$$



        or:



        $$(x_O-frac{x_A+x_B}{2})(x_B-x_A)+(y_O-frac{y_A+y_B}{2})(y_B-y_A)=0$$
        $$(x_O-frac{x_B+x_C}{2})(x_C-x_B)+(y_O-frac{y_B+y_C}{2})(y_C-y_B)=0$$



        or:



        $$(x_B-x_A)x_O+(y_B-y_A)y_O=frac{x_B^2-x_A^2+y_B^2-y_A^2}{2}tag{1}$$



        $$(x_C-x_B)x_O+(y_C-y_B)y_O=frac{x_C^2-x_B^2+y_C^2-y_B^2}{2}tag{2}$$



        The point is: equations (1) and (2) are linear wtih two unknowns $(x_O,y_O)$ and, assuming that points $A,B,C$ are not collinear, you can always find a unique solution. You can also come up with a general formula for $(x_O,y_O)$ but I leave it up to you as an exercise.



        Once you have $x_O,y_O$ you can find the radius of the circle:



        $$R=sqrt{(x_A-x_O)^2+(y_A-y_O)^2}$$



        For any point on the circle $K(x_K,y_K)$:



        $$(x_K-x_O)^2+(y_K-y_O)^2=R^2$$



        And if you know $x_K$, you can calculate the value of $y_K$:



        $$y_K=y_Opmsqrt{R^2-(x_K-x_O)^2}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 14 at 14:43









        OldboyOldboy

        8,1651936




        8,1651936






























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