Locus of Tangent point of Sphere and xy plane

kunny

In $xyz$ space with the origin $O$ , take a point $A(0,\ 1,\ 1)$ and donote by $l$ the straight line $OA$ . Suppose that the sphere $S$ with radius 1 is in the domain $z\geq 0$ and moves with touching to both the $xy$ plane and $l$ , then find the locus of the tangent point $P$ of $S$ and the $xy$ plane, then sketch the locus.

**Posted:** Sat Nov 07, 2009 7:20 am

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Today's calculation of Integral Digest

kenn4000

S can be parametrized by $1=(h(t)-x)^2+(j(t)-y)^2+(k(t)-z)^2$
with $A(t)=(h(t),j(t),k(t))$ we know $A(0)=(0,0,1)$
and we can say at time t, it is touching $(0,t,t)$ on OA, so
until time $t=2$ after which it is impossible, so $A(2)=(0,2,1)$

$1=h(t)^2+(j(t)-t)^2+(k(t)-t))^2$
to be tangent to the xy plane, we need to have only one solution $(x,y)$ to

$1=(h(t)-x)^2+(j(t)-y)^2+k(t)^2$
or, $(x-h(t))^2+(y-j(t))^2=1-k(t)^2$ so $k(t)=1$ for all t
now we get

$h(t)^2+(j(t)-t)^2=1-(1-t)^2=t(2-t)$
which is satisfied by $j(t)=t, h(t)=\sqrt{t(2-t)}$
since the center moves here, the xy tangents move along the same path with z=0 instead of 1,
$x(t)=\sqrt{t(2-t)}, y(t)=t$
which traces out on the xy plane to look like the right half of the circle $x^2+(y-1)^2=1$

**Posted:** Sat Nov 07, 2009 4:43 pm

kunny

Incorrect.

**Posted:** Mon Nov 09, 2009 5:43 pm

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atomicwedgie

I assume that line $l$ is the line passing through $(0,0,0)$ and $(0,1,1)$ , not merely the line segment $\overline{OA}$ .

Consider an equivalent situation in which a unit sphere whose center travels in the plane $y = z$ , is tangent to the $z$ -axis. Clearly, such a sphere is confined within the cylinder $x^2 + y^2 = 4$ , and within the parallel planes $|z - y| = \sqrt {2}$ . Furthermore, the center of this sphere lies on the surface of the cylinder $x^2 + y^2 = 1$ . The intersection of this cylinder with the plane $y = z$ is an ellipse, which we can parameterize as

$C(t) = (\cos\theta, \sin\theta\, \sin\theta)$ .

The locus of the point of tangency $T$ of the sphere to the lower plane $z = y - \sqrt {2}$ has the same shape as the curve $C$ , but is translated by $(0,1/\sqrt {2}, - 1/\sqrt {2})$ :

$T(t) = \left( \cos\theta, \frac {1}{\sqrt {2}} + \sin\theta, - \frac {1}{\sqrt {2}} + \sin\theta \right)$ .

The major axis of the ellipse has length $|C( - \pi/2) - C(\pi/2)| = \sqrt {2^2 + 2^2} = 2\sqrt {2}$ , and the minor axis is easily seen to be 2.

Thus, in the original coordinate system, the locus of the point of tangency is an ellipse with vertices at

$\left(0, - \cot \frac {3\pi}{8}\right) = (0, 1 - \sqrt {2})$ ,

$(0, 1 - \sqrt {2} + 2\sqrt {2}) = (0,1 + \sqrt {2})$ ,

$(1,1)$ ,

$( - 1,1)$ .

This ellipse has parametric equation

$E(t) = \left(\cos\theta, 1 + \sqrt {2} \sin\theta\right)$ .

**Posted:** Mon Nov 09, 2009 8:10 pm

kunny

That's correct, which is same as my idea.

I have another solutions in using 1. Discriminant 2. Scalar Product.

By the way, doesn't straight line $OA$ mean the line passing though the points the origin and the point $A$ ?

**Posted:** Mon Nov 09, 2009 9:54 pm

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Today's calculation of Integral Digest