Tangent to sphere in a given plane












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$begingroup$


I have a sphere centered on the origin $O$. I have a point in space $Cam$ and a vector $Dir$.



How do I find a point $X$ and a Vector Hor where:




  1. The line starting at Cam and of direction Hor, passing through X, is tangent to the sphere in X

  2. Hor, Dir and O-Cam are on the same plane


Illustration



And if possible, with the shortest calculation possible. Technically I will be implementing this in a shader, for every given direction I need to find the horizon line on the planet. I tried to abstract the problem as much as possible.






spheres







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$endgroup$

















    1












    $begingroup$


    I have a sphere centered on the origin $O$. I have a point in space $Cam$ and a vector $Dir$.



    How do I find a point $X$ and a Vector Hor where:




    1. The line starting at Cam and of direction Hor, passing through X, is tangent to the sphere in X

    2. Hor, Dir and O-Cam are on the same plane


    Illustration



    And if possible, with the shortest calculation possible. Technically I will be implementing this in a shader, for every given direction I need to find the horizon line on the planet. I tried to abstract the problem as much as possible.






    spheres







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$


      I have a sphere centered on the origin $O$. I have a point in space $Cam$ and a vector $Dir$.



      How do I find a point $X$ and a Vector Hor where:




      1. The line starting at Cam and of direction Hor, passing through X, is tangent to the sphere in X

      2. Hor, Dir and O-Cam are on the same plane


      Illustration



      And if possible, with the shortest calculation possible. Technically I will be implementing this in a shader, for every given direction I need to find the horizon line on the planet. I tried to abstract the problem as much as possible.






      spheres







      share|cite|improve this question











      $endgroup$




      I have a sphere centered on the origin $O$. I have a point in space $Cam$ and a vector $Dir$.



      How do I find a point $X$ and a Vector Hor where:




      1. The line starting at Cam and of direction Hor, passing through X, is tangent to the sphere in X

      2. Hor, Dir and O-Cam are on the same plane


      Illustration



      And if possible, with the shortest calculation possible. Technically I will be implementing this in a shader, for every given direction I need to find the horizon line on the planet. I tried to abstract the problem as much as possible.






      spheres




      spheres






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Feb 16 '16 at 19:53









      Pichi Wuana

      658525




      658525










      asked Feb 16 '16 at 19:44









      blackrackblackrack

      62




      62






















          2 Answers
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          active

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          0












          $begingroup$

          You stated that the center of the sphere is the origin of our 3-dimensional space, so suppose we have $O = (0, 0, 0)$, $Cam=(c_x, c_y, c_z)$ and $overrightarrow{Dir}=langle d_x, d_y, d_z rangle$, and that the radius of the sphere is $R$.



          Let $X = (r_x, r_y, r_z)$ be what we are solving for.



          First, we need to find the plane containing both $Dir$ and $Cam$. Taking the cross product of $overrightarrow{Dir}$ and $overrightarrow{CamO}$, combined with the fact that the plane should contain $O = (0, 0, 0)$, the equation of our plane $P$ is $(c_zd_y - c_yd_z)x + (c_xd_z - c_zd_x)y + (c_yd_x - c_xd_y)z = 0$. We will later use this to ensure our point $X$ is on this plane.



          Next, consider the conditions for $X$ to be the point at which $overrightarrow{Hor}$ tangentially contacts the sphere. We need the point $X$ to be on the surface of the sphere, and we need the vectors $overrightarrow{XO}$ and $overrightarrow{XCam}$ to be perpendicular (in order for $overrightarrow{XCam}$ to be tangent to the sphere).



          For the first condition, we want $r_x^2 + r_y^2 + r_z^2 = R^2$, simply by substituting the coordinates of $X$ into the equation of our sphere.



          For the second condition, we want $overrightarrow{XO} cdot overrightarrow{XCam} = 0$. Since $overrightarrow{XO} = langle -r_x, -r_y, -r_z rangle$ and $overrightarrow{XCam} = langle c_x - r_x, c_y - r_y, c_z - r_z rangle$, we have $overrightarrow{XO} cdot overrightarrow{XCam} = r_x^2 - c_xr_x + r_y^2 - c_yr_y + r_z^2 - c_zr_z = 0$. Note that we can simplify this equation by first substituting $r_x^2 + r_y^2 + r_z^2 = R^2$, and we obtain $c_xr_x + c_yr_y + c_zr_z = R^2$.



          Thus, we now have a system of 3 equations and 3 unknowns:




          • $(c_zd_y - c_yd_z)r_x + (c_xd_z - c_zd_x)r_y + (c_yd_x - c_xd_y)r_z = 0$

          • $r_x^2 + r_y^2 + r_z^2 = R^2$

          • $c_xr_x + c_yr_y + c_zr_z = R^2$


          Now you can just pick your favorite way to solve it. This will give you the coordinates of $X$, and you will in turn have $overrightarrow{Hor} = overrightarrow{CamX}$. I haven't done the actual substitution yet as it's a huge mess, but I think this answers your question. I'm not exactly sure what you mean by the shortest calculation possible, but I think this method is straightforward and gives you a general expression for the coordinates of $X$ in terms of the parameters you provided.



          Hope this helped.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks a lot for your response. I tried to solve that system but after a bit of frustration I gave up on it and reformulated the problem as follows: 1) The problem could be assimilated to an intersection of two circles, since we're working in a plane. Once circle centered on the sphere center of radius r, and a second circle centered on Cam of radius (Cam-X) 2) The length of (Cam-X) can be determined from (Cam-O) and the radius of the sphere, taking advantage of the fact that (O-X) and (Cam-X) are perpendicular since X is a tangent. The resulting problem was much easier for me to solve.
            $endgroup$
            – blackrack
            Feb 17 '16 at 3:13












          • $begingroup$
            @blackrack No problem. Indeed, that sounds like a nice way to do it. Was it really easier to solve? I'm envisioning solving for the intersection of two spheres and a plane (also a system of 3 equations), which to me feels like would still take a lot of work to grind through. How did you go about solving it?
            $endgroup$
            – Brian Yao
            Feb 17 '16 at 4:48










          • $begingroup$
            I actually solved for the intersection of 2 circles and used existing circle intersection formulas instead of intersection of 2 spheres and a plane, I'll describe and illustrate my idea shortly.
            $endgroup$
            – blackrack
            Feb 17 '16 at 14:33



















          0












          $begingroup$

          Create a local coordinate system on the plane with the center of the circle as origin, and local x direction along $vec{OC_{am}}$. The 3×3 rotation matrix is
          $$ begin{align}
          E = & begin{vmatrix} hat{i} & hat{j} & hat{k} end{vmatrix} \
          hat{i} &= [ vec{O C_{am}} ] \
          hat{k} & = [hat{i} times vec{Dir}] \
          hat{j} &= hat{k} times hat{i}
          end{align} $$



          where the $[vec{v}] = frac{vec{v}}{| vec{v} |}$ notation is for unit vectors, and $times$ is the vector cross product.



          Now you take the distance between the center of the circle and point $O_{am}$ along the $hat{i}$ direction as $ell = | vec{O C_{am} } |$



          The $(x,y)$ coordinate of the tangent point X on the plane are



          $$ (x,y) = left( frac{r^2}{ell}, frac{r sqrt{ell^2-r^2}}{ell} right) $$
          where $r$ is the radius of the circle.



          The 3D coordinates of X are



          $$ vec{X} = frac{r^2}{ell} hat{i} + frac{r sqrt{ell^2-r^2}}{ell} hat{j} $$



          The direction of X is $$ vec{H_{or}} = [vec{O_{am} X}] $$



          NOTE: The equation of the tangent line on the plane is $left(frac{r}{ell}right)x + left( frac{sqrt{ell^2-r^2}}{ell} right)y = r$.



          Example



          I have verified the calculation with a GeoGebra model:



          geo






          share|cite|improve this answer











          $endgroup$














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            2 Answers
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            2 Answers
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            0












            $begingroup$

            You stated that the center of the sphere is the origin of our 3-dimensional space, so suppose we have $O = (0, 0, 0)$, $Cam=(c_x, c_y, c_z)$ and $overrightarrow{Dir}=langle d_x, d_y, d_z rangle$, and that the radius of the sphere is $R$.



            Let $X = (r_x, r_y, r_z)$ be what we are solving for.



            First, we need to find the plane containing both $Dir$ and $Cam$. Taking the cross product of $overrightarrow{Dir}$ and $overrightarrow{CamO}$, combined with the fact that the plane should contain $O = (0, 0, 0)$, the equation of our plane $P$ is $(c_zd_y - c_yd_z)x + (c_xd_z - c_zd_x)y + (c_yd_x - c_xd_y)z = 0$. We will later use this to ensure our point $X$ is on this plane.



            Next, consider the conditions for $X$ to be the point at which $overrightarrow{Hor}$ tangentially contacts the sphere. We need the point $X$ to be on the surface of the sphere, and we need the vectors $overrightarrow{XO}$ and $overrightarrow{XCam}$ to be perpendicular (in order for $overrightarrow{XCam}$ to be tangent to the sphere).



            For the first condition, we want $r_x^2 + r_y^2 + r_z^2 = R^2$, simply by substituting the coordinates of $X$ into the equation of our sphere.



            For the second condition, we want $overrightarrow{XO} cdot overrightarrow{XCam} = 0$. Since $overrightarrow{XO} = langle -r_x, -r_y, -r_z rangle$ and $overrightarrow{XCam} = langle c_x - r_x, c_y - r_y, c_z - r_z rangle$, we have $overrightarrow{XO} cdot overrightarrow{XCam} = r_x^2 - c_xr_x + r_y^2 - c_yr_y + r_z^2 - c_zr_z = 0$. Note that we can simplify this equation by first substituting $r_x^2 + r_y^2 + r_z^2 = R^2$, and we obtain $c_xr_x + c_yr_y + c_zr_z = R^2$.



            Thus, we now have a system of 3 equations and 3 unknowns:




            • $(c_zd_y - c_yd_z)r_x + (c_xd_z - c_zd_x)r_y + (c_yd_x - c_xd_y)r_z = 0$

            • $r_x^2 + r_y^2 + r_z^2 = R^2$

            • $c_xr_x + c_yr_y + c_zr_z = R^2$


            Now you can just pick your favorite way to solve it. This will give you the coordinates of $X$, and you will in turn have $overrightarrow{Hor} = overrightarrow{CamX}$. I haven't done the actual substitution yet as it's a huge mess, but I think this answers your question. I'm not exactly sure what you mean by the shortest calculation possible, but I think this method is straightforward and gives you a general expression for the coordinates of $X$ in terms of the parameters you provided.



            Hope this helped.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thanks a lot for your response. I tried to solve that system but after a bit of frustration I gave up on it and reformulated the problem as follows: 1) The problem could be assimilated to an intersection of two circles, since we're working in a plane. Once circle centered on the sphere center of radius r, and a second circle centered on Cam of radius (Cam-X) 2) The length of (Cam-X) can be determined from (Cam-O) and the radius of the sphere, taking advantage of the fact that (O-X) and (Cam-X) are perpendicular since X is a tangent. The resulting problem was much easier for me to solve.
              $endgroup$
              – blackrack
              Feb 17 '16 at 3:13












            • $begingroup$
              @blackrack No problem. Indeed, that sounds like a nice way to do it. Was it really easier to solve? I'm envisioning solving for the intersection of two spheres and a plane (also a system of 3 equations), which to me feels like would still take a lot of work to grind through. How did you go about solving it?
              $endgroup$
              – Brian Yao
              Feb 17 '16 at 4:48










            • $begingroup$
              I actually solved for the intersection of 2 circles and used existing circle intersection formulas instead of intersection of 2 spheres and a plane, I'll describe and illustrate my idea shortly.
              $endgroup$
              – blackrack
              Feb 17 '16 at 14:33
















            0












            $begingroup$

            You stated that the center of the sphere is the origin of our 3-dimensional space, so suppose we have $O = (0, 0, 0)$, $Cam=(c_x, c_y, c_z)$ and $overrightarrow{Dir}=langle d_x, d_y, d_z rangle$, and that the radius of the sphere is $R$.



            Let $X = (r_x, r_y, r_z)$ be what we are solving for.



            First, we need to find the plane containing both $Dir$ and $Cam$. Taking the cross product of $overrightarrow{Dir}$ and $overrightarrow{CamO}$, combined with the fact that the plane should contain $O = (0, 0, 0)$, the equation of our plane $P$ is $(c_zd_y - c_yd_z)x + (c_xd_z - c_zd_x)y + (c_yd_x - c_xd_y)z = 0$. We will later use this to ensure our point $X$ is on this plane.



            Next, consider the conditions for $X$ to be the point at which $overrightarrow{Hor}$ tangentially contacts the sphere. We need the point $X$ to be on the surface of the sphere, and we need the vectors $overrightarrow{XO}$ and $overrightarrow{XCam}$ to be perpendicular (in order for $overrightarrow{XCam}$ to be tangent to the sphere).



            For the first condition, we want $r_x^2 + r_y^2 + r_z^2 = R^2$, simply by substituting the coordinates of $X$ into the equation of our sphere.



            For the second condition, we want $overrightarrow{XO} cdot overrightarrow{XCam} = 0$. Since $overrightarrow{XO} = langle -r_x, -r_y, -r_z rangle$ and $overrightarrow{XCam} = langle c_x - r_x, c_y - r_y, c_z - r_z rangle$, we have $overrightarrow{XO} cdot overrightarrow{XCam} = r_x^2 - c_xr_x + r_y^2 - c_yr_y + r_z^2 - c_zr_z = 0$. Note that we can simplify this equation by first substituting $r_x^2 + r_y^2 + r_z^2 = R^2$, and we obtain $c_xr_x + c_yr_y + c_zr_z = R^2$.



            Thus, we now have a system of 3 equations and 3 unknowns:




            • $(c_zd_y - c_yd_z)r_x + (c_xd_z - c_zd_x)r_y + (c_yd_x - c_xd_y)r_z = 0$

            • $r_x^2 + r_y^2 + r_z^2 = R^2$

            • $c_xr_x + c_yr_y + c_zr_z = R^2$


            Now you can just pick your favorite way to solve it. This will give you the coordinates of $X$, and you will in turn have $overrightarrow{Hor} = overrightarrow{CamX}$. I haven't done the actual substitution yet as it's a huge mess, but I think this answers your question. I'm not exactly sure what you mean by the shortest calculation possible, but I think this method is straightforward and gives you a general expression for the coordinates of $X$ in terms of the parameters you provided.



            Hope this helped.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thanks a lot for your response. I tried to solve that system but after a bit of frustration I gave up on it and reformulated the problem as follows: 1) The problem could be assimilated to an intersection of two circles, since we're working in a plane. Once circle centered on the sphere center of radius r, and a second circle centered on Cam of radius (Cam-X) 2) The length of (Cam-X) can be determined from (Cam-O) and the radius of the sphere, taking advantage of the fact that (O-X) and (Cam-X) are perpendicular since X is a tangent. The resulting problem was much easier for me to solve.
              $endgroup$
              – blackrack
              Feb 17 '16 at 3:13












            • $begingroup$
              @blackrack No problem. Indeed, that sounds like a nice way to do it. Was it really easier to solve? I'm envisioning solving for the intersection of two spheres and a plane (also a system of 3 equations), which to me feels like would still take a lot of work to grind through. How did you go about solving it?
              $endgroup$
              – Brian Yao
              Feb 17 '16 at 4:48










            • $begingroup$
              I actually solved for the intersection of 2 circles and used existing circle intersection formulas instead of intersection of 2 spheres and a plane, I'll describe and illustrate my idea shortly.
              $endgroup$
              – blackrack
              Feb 17 '16 at 14:33














            0












            0








            0





            $begingroup$

            You stated that the center of the sphere is the origin of our 3-dimensional space, so suppose we have $O = (0, 0, 0)$, $Cam=(c_x, c_y, c_z)$ and $overrightarrow{Dir}=langle d_x, d_y, d_z rangle$, and that the radius of the sphere is $R$.



            Let $X = (r_x, r_y, r_z)$ be what we are solving for.



            First, we need to find the plane containing both $Dir$ and $Cam$. Taking the cross product of $overrightarrow{Dir}$ and $overrightarrow{CamO}$, combined with the fact that the plane should contain $O = (0, 0, 0)$, the equation of our plane $P$ is $(c_zd_y - c_yd_z)x + (c_xd_z - c_zd_x)y + (c_yd_x - c_xd_y)z = 0$. We will later use this to ensure our point $X$ is on this plane.



            Next, consider the conditions for $X$ to be the point at which $overrightarrow{Hor}$ tangentially contacts the sphere. We need the point $X$ to be on the surface of the sphere, and we need the vectors $overrightarrow{XO}$ and $overrightarrow{XCam}$ to be perpendicular (in order for $overrightarrow{XCam}$ to be tangent to the sphere).



            For the first condition, we want $r_x^2 + r_y^2 + r_z^2 = R^2$, simply by substituting the coordinates of $X$ into the equation of our sphere.



            For the second condition, we want $overrightarrow{XO} cdot overrightarrow{XCam} = 0$. Since $overrightarrow{XO} = langle -r_x, -r_y, -r_z rangle$ and $overrightarrow{XCam} = langle c_x - r_x, c_y - r_y, c_z - r_z rangle$, we have $overrightarrow{XO} cdot overrightarrow{XCam} = r_x^2 - c_xr_x + r_y^2 - c_yr_y + r_z^2 - c_zr_z = 0$. Note that we can simplify this equation by first substituting $r_x^2 + r_y^2 + r_z^2 = R^2$, and we obtain $c_xr_x + c_yr_y + c_zr_z = R^2$.



            Thus, we now have a system of 3 equations and 3 unknowns:




            • $(c_zd_y - c_yd_z)r_x + (c_xd_z - c_zd_x)r_y + (c_yd_x - c_xd_y)r_z = 0$

            • $r_x^2 + r_y^2 + r_z^2 = R^2$

            • $c_xr_x + c_yr_y + c_zr_z = R^2$


            Now you can just pick your favorite way to solve it. This will give you the coordinates of $X$, and you will in turn have $overrightarrow{Hor} = overrightarrow{CamX}$. I haven't done the actual substitution yet as it's a huge mess, but I think this answers your question. I'm not exactly sure what you mean by the shortest calculation possible, but I think this method is straightforward and gives you a general expression for the coordinates of $X$ in terms of the parameters you provided.



            Hope this helped.






            share|cite|improve this answer









            $endgroup$



            You stated that the center of the sphere is the origin of our 3-dimensional space, so suppose we have $O = (0, 0, 0)$, $Cam=(c_x, c_y, c_z)$ and $overrightarrow{Dir}=langle d_x, d_y, d_z rangle$, and that the radius of the sphere is $R$.



            Let $X = (r_x, r_y, r_z)$ be what we are solving for.



            First, we need to find the plane containing both $Dir$ and $Cam$. Taking the cross product of $overrightarrow{Dir}$ and $overrightarrow{CamO}$, combined with the fact that the plane should contain $O = (0, 0, 0)$, the equation of our plane $P$ is $(c_zd_y - c_yd_z)x + (c_xd_z - c_zd_x)y + (c_yd_x - c_xd_y)z = 0$. We will later use this to ensure our point $X$ is on this plane.



            Next, consider the conditions for $X$ to be the point at which $overrightarrow{Hor}$ tangentially contacts the sphere. We need the point $X$ to be on the surface of the sphere, and we need the vectors $overrightarrow{XO}$ and $overrightarrow{XCam}$ to be perpendicular (in order for $overrightarrow{XCam}$ to be tangent to the sphere).



            For the first condition, we want $r_x^2 + r_y^2 + r_z^2 = R^2$, simply by substituting the coordinates of $X$ into the equation of our sphere.



            For the second condition, we want $overrightarrow{XO} cdot overrightarrow{XCam} = 0$. Since $overrightarrow{XO} = langle -r_x, -r_y, -r_z rangle$ and $overrightarrow{XCam} = langle c_x - r_x, c_y - r_y, c_z - r_z rangle$, we have $overrightarrow{XO} cdot overrightarrow{XCam} = r_x^2 - c_xr_x + r_y^2 - c_yr_y + r_z^2 - c_zr_z = 0$. Note that we can simplify this equation by first substituting $r_x^2 + r_y^2 + r_z^2 = R^2$, and we obtain $c_xr_x + c_yr_y + c_zr_z = R^2$.



            Thus, we now have a system of 3 equations and 3 unknowns:




            • $(c_zd_y - c_yd_z)r_x + (c_xd_z - c_zd_x)r_y + (c_yd_x - c_xd_y)r_z = 0$

            • $r_x^2 + r_y^2 + r_z^2 = R^2$

            • $c_xr_x + c_yr_y + c_zr_z = R^2$


            Now you can just pick your favorite way to solve it. This will give you the coordinates of $X$, and you will in turn have $overrightarrow{Hor} = overrightarrow{CamX}$. I haven't done the actual substitution yet as it's a huge mess, but I think this answers your question. I'm not exactly sure what you mean by the shortest calculation possible, but I think this method is straightforward and gives you a general expression for the coordinates of $X$ in terms of the parameters you provided.



            Hope this helped.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Feb 16 '16 at 22:11









            Brian YaoBrian Yao

            533




            533












            • $begingroup$
              Thanks a lot for your response. I tried to solve that system but after a bit of frustration I gave up on it and reformulated the problem as follows: 1) The problem could be assimilated to an intersection of two circles, since we're working in a plane. Once circle centered on the sphere center of radius r, and a second circle centered on Cam of radius (Cam-X) 2) The length of (Cam-X) can be determined from (Cam-O) and the radius of the sphere, taking advantage of the fact that (O-X) and (Cam-X) are perpendicular since X is a tangent. The resulting problem was much easier for me to solve.
              $endgroup$
              – blackrack
              Feb 17 '16 at 3:13












            • $begingroup$
              @blackrack No problem. Indeed, that sounds like a nice way to do it. Was it really easier to solve? I'm envisioning solving for the intersection of two spheres and a plane (also a system of 3 equations), which to me feels like would still take a lot of work to grind through. How did you go about solving it?
              $endgroup$
              – Brian Yao
              Feb 17 '16 at 4:48










            • $begingroup$
              I actually solved for the intersection of 2 circles and used existing circle intersection formulas instead of intersection of 2 spheres and a plane, I'll describe and illustrate my idea shortly.
              $endgroup$
              – blackrack
              Feb 17 '16 at 14:33


















            • $begingroup$
              Thanks a lot for your response. I tried to solve that system but after a bit of frustration I gave up on it and reformulated the problem as follows: 1) The problem could be assimilated to an intersection of two circles, since we're working in a plane. Once circle centered on the sphere center of radius r, and a second circle centered on Cam of radius (Cam-X) 2) The length of (Cam-X) can be determined from (Cam-O) and the radius of the sphere, taking advantage of the fact that (O-X) and (Cam-X) are perpendicular since X is a tangent. The resulting problem was much easier for me to solve.
              $endgroup$
              – blackrack
              Feb 17 '16 at 3:13












            • $begingroup$
              @blackrack No problem. Indeed, that sounds like a nice way to do it. Was it really easier to solve? I'm envisioning solving for the intersection of two spheres and a plane (also a system of 3 equations), which to me feels like would still take a lot of work to grind through. How did you go about solving it?
              $endgroup$
              – Brian Yao
              Feb 17 '16 at 4:48










            • $begingroup$
              I actually solved for the intersection of 2 circles and used existing circle intersection formulas instead of intersection of 2 spheres and a plane, I'll describe and illustrate my idea shortly.
              $endgroup$
              – blackrack
              Feb 17 '16 at 14:33
















            $begingroup$
            Thanks a lot for your response. I tried to solve that system but after a bit of frustration I gave up on it and reformulated the problem as follows: 1) The problem could be assimilated to an intersection of two circles, since we're working in a plane. Once circle centered on the sphere center of radius r, and a second circle centered on Cam of radius (Cam-X) 2) The length of (Cam-X) can be determined from (Cam-O) and the radius of the sphere, taking advantage of the fact that (O-X) and (Cam-X) are perpendicular since X is a tangent. The resulting problem was much easier for me to solve.
            $endgroup$
            – blackrack
            Feb 17 '16 at 3:13






            $begingroup$
            Thanks a lot for your response. I tried to solve that system but after a bit of frustration I gave up on it and reformulated the problem as follows: 1) The problem could be assimilated to an intersection of two circles, since we're working in a plane. Once circle centered on the sphere center of radius r, and a second circle centered on Cam of radius (Cam-X) 2) The length of (Cam-X) can be determined from (Cam-O) and the radius of the sphere, taking advantage of the fact that (O-X) and (Cam-X) are perpendicular since X is a tangent. The resulting problem was much easier for me to solve.
            $endgroup$
            – blackrack
            Feb 17 '16 at 3:13














            $begingroup$
            @blackrack No problem. Indeed, that sounds like a nice way to do it. Was it really easier to solve? I'm envisioning solving for the intersection of two spheres and a plane (also a system of 3 equations), which to me feels like would still take a lot of work to grind through. How did you go about solving it?
            $endgroup$
            – Brian Yao
            Feb 17 '16 at 4:48




            $begingroup$
            @blackrack No problem. Indeed, that sounds like a nice way to do it. Was it really easier to solve? I'm envisioning solving for the intersection of two spheres and a plane (also a system of 3 equations), which to me feels like would still take a lot of work to grind through. How did you go about solving it?
            $endgroup$
            – Brian Yao
            Feb 17 '16 at 4:48












            $begingroup$
            I actually solved for the intersection of 2 circles and used existing circle intersection formulas instead of intersection of 2 spheres and a plane, I'll describe and illustrate my idea shortly.
            $endgroup$
            – blackrack
            Feb 17 '16 at 14:33




            $begingroup$
            I actually solved for the intersection of 2 circles and used existing circle intersection formulas instead of intersection of 2 spheres and a plane, I'll describe and illustrate my idea shortly.
            $endgroup$
            – blackrack
            Feb 17 '16 at 14:33











            0












            $begingroup$

            Create a local coordinate system on the plane with the center of the circle as origin, and local x direction along $vec{OC_{am}}$. The 3×3 rotation matrix is
            $$ begin{align}
            E = & begin{vmatrix} hat{i} & hat{j} & hat{k} end{vmatrix} \
            hat{i} &= [ vec{O C_{am}} ] \
            hat{k} & = [hat{i} times vec{Dir}] \
            hat{j} &= hat{k} times hat{i}
            end{align} $$



            where the $[vec{v}] = frac{vec{v}}{| vec{v} |}$ notation is for unit vectors, and $times$ is the vector cross product.



            Now you take the distance between the center of the circle and point $O_{am}$ along the $hat{i}$ direction as $ell = | vec{O C_{am} } |$



            The $(x,y)$ coordinate of the tangent point X on the plane are



            $$ (x,y) = left( frac{r^2}{ell}, frac{r sqrt{ell^2-r^2}}{ell} right) $$
            where $r$ is the radius of the circle.



            The 3D coordinates of X are



            $$ vec{X} = frac{r^2}{ell} hat{i} + frac{r sqrt{ell^2-r^2}}{ell} hat{j} $$



            The direction of X is $$ vec{H_{or}} = [vec{O_{am} X}] $$



            NOTE: The equation of the tangent line on the plane is $left(frac{r}{ell}right)x + left( frac{sqrt{ell^2-r^2}}{ell} right)y = r$.



            Example



            I have verified the calculation with a GeoGebra model:



            geo






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              Create a local coordinate system on the plane with the center of the circle as origin, and local x direction along $vec{OC_{am}}$. The 3×3 rotation matrix is
              $$ begin{align}
              E = & begin{vmatrix} hat{i} & hat{j} & hat{k} end{vmatrix} \
              hat{i} &= [ vec{O C_{am}} ] \
              hat{k} & = [hat{i} times vec{Dir}] \
              hat{j} &= hat{k} times hat{i}
              end{align} $$



              where the $[vec{v}] = frac{vec{v}}{| vec{v} |}$ notation is for unit vectors, and $times$ is the vector cross product.



              Now you take the distance between the center of the circle and point $O_{am}$ along the $hat{i}$ direction as $ell = | vec{O C_{am} } |$



              The $(x,y)$ coordinate of the tangent point X on the plane are



              $$ (x,y) = left( frac{r^2}{ell}, frac{r sqrt{ell^2-r^2}}{ell} right) $$
              where $r$ is the radius of the circle.



              The 3D coordinates of X are



              $$ vec{X} = frac{r^2}{ell} hat{i} + frac{r sqrt{ell^2-r^2}}{ell} hat{j} $$



              The direction of X is $$ vec{H_{or}} = [vec{O_{am} X}] $$



              NOTE: The equation of the tangent line on the plane is $left(frac{r}{ell}right)x + left( frac{sqrt{ell^2-r^2}}{ell} right)y = r$.



              Example



              I have verified the calculation with a GeoGebra model:



              geo






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                Create a local coordinate system on the plane with the center of the circle as origin, and local x direction along $vec{OC_{am}}$. The 3×3 rotation matrix is
                $$ begin{align}
                E = & begin{vmatrix} hat{i} & hat{j} & hat{k} end{vmatrix} \
                hat{i} &= [ vec{O C_{am}} ] \
                hat{k} & = [hat{i} times vec{Dir}] \
                hat{j} &= hat{k} times hat{i}
                end{align} $$



                where the $[vec{v}] = frac{vec{v}}{| vec{v} |}$ notation is for unit vectors, and $times$ is the vector cross product.



                Now you take the distance between the center of the circle and point $O_{am}$ along the $hat{i}$ direction as $ell = | vec{O C_{am} } |$



                The $(x,y)$ coordinate of the tangent point X on the plane are



                $$ (x,y) = left( frac{r^2}{ell}, frac{r sqrt{ell^2-r^2}}{ell} right) $$
                where $r$ is the radius of the circle.



                The 3D coordinates of X are



                $$ vec{X} = frac{r^2}{ell} hat{i} + frac{r sqrt{ell^2-r^2}}{ell} hat{j} $$



                The direction of X is $$ vec{H_{or}} = [vec{O_{am} X}] $$



                NOTE: The equation of the tangent line on the plane is $left(frac{r}{ell}right)x + left( frac{sqrt{ell^2-r^2}}{ell} right)y = r$.



                Example



                I have verified the calculation with a GeoGebra model:



                geo






                share|cite|improve this answer











                $endgroup$



                Create a local coordinate system on the plane with the center of the circle as origin, and local x direction along $vec{OC_{am}}$. The 3×3 rotation matrix is
                $$ begin{align}
                E = & begin{vmatrix} hat{i} & hat{j} & hat{k} end{vmatrix} \
                hat{i} &= [ vec{O C_{am}} ] \
                hat{k} & = [hat{i} times vec{Dir}] \
                hat{j} &= hat{k} times hat{i}
                end{align} $$



                where the $[vec{v}] = frac{vec{v}}{| vec{v} |}$ notation is for unit vectors, and $times$ is the vector cross product.



                Now you take the distance between the center of the circle and point $O_{am}$ along the $hat{i}$ direction as $ell = | vec{O C_{am} } |$



                The $(x,y)$ coordinate of the tangent point X on the plane are



                $$ (x,y) = left( frac{r^2}{ell}, frac{r sqrt{ell^2-r^2}}{ell} right) $$
                where $r$ is the radius of the circle.



                The 3D coordinates of X are



                $$ vec{X} = frac{r^2}{ell} hat{i} + frac{r sqrt{ell^2-r^2}}{ell} hat{j} $$



                The direction of X is $$ vec{H_{or}} = [vec{O_{am} X}] $$



                NOTE: The equation of the tangent line on the plane is $left(frac{r}{ell}right)x + left( frac{sqrt{ell^2-r^2}}{ell} right)y = r$.



                Example



                I have verified the calculation with a GeoGebra model:



                geo







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Feb 22 '16 at 16:49

























                answered Feb 22 '16 at 15:44









                ja72ja72

                7,54212044




                7,54212044






























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