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I would like to know parametric equation of the tangent point of a sphere and a circle. Circle center point is $(a,b,c)$ and its radius is $K$. Sphere center point is $(d,e,f)$ and its radius is $L$.

I assume that specified circle and a sphere intersect on one tangent point. I would like to know how I can determine the tangent point $(x,y,z)$.

If there is a misinformation please define for me.

Thanks in advance.

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  • $\begingroup$ Are you wondering about the determination of a circle tangent to a sphere given its respective centers, the plane of the circle and the ray of the sphere? $\endgroup$
    – Cesareo
    Aug 7, 2019 at 7:24
  • $\begingroup$ This is equivalent to finding the intersection of two spheres. $\endgroup$
    – amd
    Aug 7, 2019 at 8:36
  • $\begingroup$ I exactly want to find the cartesian coordinate of the tangent point regarding to given center point and radius of sphere and circle. $\endgroup$
    – Onur
    Aug 7, 2019 at 9:45
  • $\begingroup$ A circle outside a sphere and touching it could be rotated about the axis between their centers and trace a circle on the sphere orthogonal to that axis. Is that what you want? As @amd says, this is the circle of intersection of 2 spheres. $\endgroup$
    – Paul
    Aug 7, 2019 at 11:23
  • $\begingroup$ Actually, I think that the question can be asked for intersection point of two spheres. Because circle and sphere have same equations. But I would like to know one specific point that they are tangent. Because if the spheres intersect with each other, it will have two intersection point. But I need only one tangent point. $\endgroup$
    – Onur
    Aug 8, 2019 at 4:50

1 Answer 1

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In the plane passing through the centers $A,B$ and the tangent point $C$, you get a triangle $ABC$ of which you know the three sides.
So you can determine every parameter of it, and in particular the height $h$ from $C$ to the side $AB$.

When reported in $3$D, the triangle is free to rotate around the axis $AB$ and the point C can be whichever along a circle on the sphere of radius $h$ from the $AB$ axis.

You need one more constraint to fix the circle.

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  • $\begingroup$ I know the plane equation of the circle that is lying on. On the other hand, the two intersection points can be useful for me. $\endgroup$
    – Onur
    Aug 8, 2019 at 5:56
  • $\begingroup$ @onur: if the plane of the circle is known, then the plane $ABC$ is also known as being normal to the first. Then the problem is much easier: please specify exactly what is known and what is to be determined. $\endgroup$
    – G Cab
    Aug 8, 2019 at 14:20

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