Tangent Planes

**JoAdams5000** · July 3rd 2009, 08:22 AM

I am trying to find all planes that are tangent to both x^2 + y^2 + z^2 = 1 and x^2 + y^2 + 2z = 0.

I tried using gradient to get the general equation of the planes for each but after that I got stuck.

Any suggestions would be greatly appreciated. Thanks!

**halbard** · July 3rd 2009, 09:39 AM

Here's one way.

The general tangent plane to the sphere at the point $(x_1,y_1,z_1)$ is $xx_1+yy_1+zz_1-1=0$ , and the general tangent plane to the paraboloid at $(x_2,y_2,z_2)$ is $xx_2+yy_2+z+z_2=0$ .

If these planes are identical, we must have $\frac{x_1}{x_2}=\frac{y_1}{y_2}=\frac{z_1}1=\frac{-1}{z_2}$ .

So $x_1=-\frac{x_2}{z_2}$ , $y_1=-\frac{y_2}{z_2}$ and $z_1=-\frac1{z_2}$ . Using $x_1^2+y_1^2+z_1^2=1$ we deduce that $x_2^2+y_2^2+1=z_2^2$ .

But $x_2^2+y_2^2=-2z_2$ . Therefore $z_2^2+2z_2-1=0$ and $z_2<0$ implying $z_2=-1-\surd2$ .

Hence $z_1=-\frac1{z_2}=-1+\surd2$ , $x_1^2+y_1^2=1-z_1^2=2(-1+\surd 2)$ and $x_2^2+y_2^2=-2z_2=2(1+\surd 2)$ .

To summarise: with $k=1+\surd 2$ the common tangent planes are $xx_2+yy_2+z=k$ where $x_2^2+y_2^2=2k$ , touching the sphere at $(x_2/k,y_2/k,1/k)$ and the paraboloid at $(x_2,y_2,-k)$ .

**simplependulum** · July 4th 2009, 08:09 PM

$x^2 + y^2 + z^2 = 1~~$ (1)
$x^2 + y^2 + 2x = 0~~$ (2)

Use a parametric equation to replace (2) ,

$x = r \cos{ \theta} , y = r \sin{\theta} , z = - \frac{r^2}{2}$

And the normal of the paraboloid ( in vector form , but not unit vector) is :

$\nabla (x^2 + y^2 + 2z) = 2x i + 2y j + 2k = 2r\cos{\theta} i + 2r\sin{\theta}j + 2k$

And the tangent plane of it is :

$2(x-r \cos{ \theta})(r \cos{ \theta}) + 2(y-r \sin{\theta})(r \sin{\theta}) + 2z + r^2 = 0$

OR

$r \cos{ \theta} x + r \sin{ \theta} y + z - \frac{r^2}{2} = 0$

Since the required planes touch the unit sphere , the distance of the plane from the oringin is 1 .

$| \frac{r \cos{ \theta} (0) + r \sin{ \theta} (0) + (0) - \frac{r^2}{2}}{ \sqrt{ r^2 + 1} } | = 1$

$r^4 - 4r^2 - 4 = 0$

Solving the equation , gives $r = \sqrt{2 + 2\sqrt{2}}$

Finally , substitute back to the equation of the plane ,

the required plane is obtained :

$r \cos{ \theta} x + r \sin{ \theta} y + z - \frac{r^2}{2} = 0$

In fact , there are infinite planes which satisify your requirement

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