# Find the point of contact of the plane $2x-2y+z+12=0$ with the sphere $x^2+y^2+z^2-2x-4y+2z=3.$

Find the point of contact of the plane $$2x-2y+z+12=0$$ with the sphere $$x^2+y^2+z^2-2x-4y+2z=3.$$

I already know the solution of this problem. However, while solving this, I first tried solving it in the following approach:

We know that the equation of a tangent plane of the sphere $$x^2+y^2+z^2+2ux+2vy+2wz+d=0$$(,where $$(-u,-v,-w)$$ is the centre of the sphere) at the point $$(x_1,y_1,z_1)$$ is $$xx_1+yy_1+zz_1+(x+x_1)u+(y+y_1)v+(z+z_1)w+d=0.$$ Given, the equation of the plane $$2x-2y+z+12=0$$ and comparing it with $$xx_1+yy_1+zz_1+(x+x_1)u+(y+y_1)v+(z+z_1)w+d=x(x_1+u)+y(y_1+v)+z(z_1+w)+x_1u+y_1v+z_1w+d=0,$$ where $$(u,v,w)=(-1,2,1)$$ we get, $$u+x_1=2,v+y_1=-2,w+z_1=1$$ and hence, $$x_1=3,y_1=0,z_1=0$$ due to which $$(3,0,0)$$ is the point of contact.

Now, this solution is obviously wrong as the correct point of contact will be $$(-1,4,-2).$$ But I don’t understand what's going wrong with this approach? To be specific: I want to know, why we can't apply this approach ?

• I think you forgot that the plane equation can be scaled. You need to match the scale of the two equations (via the $d$ term) first, then set the $x,y,z$ coefficients equal Feb 6 at 18:54
• (actually, not sure that exact approach will work, but I think that's the flaw -- you need 4 equations, not 3) Feb 6 at 18:56
• Check and you will find that $y_1=0$, not $4$. The point $(3,0,0)$ is the other end of the diameter of the sphere containing the point $(-1,4,-2)$ on its surface. Feb 6 at 19:51
• @JohnWaylandBales Thanks for pointing out the typo! I have edited it. I think the most obvious reason that this approach is wrong because $(3,0,0)$ doesn't actually lie on the tangent plane. Thus, by this approach we are not getting the tangent point.This might br the true reason. What do you think? Feb 7 at 4:36
• @AlexK Actually, it's true, that I did not match the $d$ term and in my case $(3,0,0)$, the d-term is not matching at all. But what bothers me, is that, why is it happening so, i.e why the $d-term$ is not matching in both the cases since, they are both equations of tangent plane and comparing them the $d$ term should've matched? Feb 7 at 6:03

You can easily check that $$(-3,4,0)$$ is not on the plane given by $$2x-2y+z+12=0$$ because $$2(-3)-2(4)+0+12=-14+12=-2\ne 0$$. I don't understand what your solution is trying to do because you haven't explained what $$u$$, $$v$$ and $$w$$ are. Adding unknowns is usually not the way to go.

An easier way to solve this is to consider the wording. We are looking for a single point of contact which means that the plane is tangent to the sphere which means that the vector that is normal to the plane is parallel to the gradient of a function that has the sphere as a level surface.

Let $$f(x,y,z)=x^2+y^2+z^2-2x-4y+2z$$ Then the sphere is the set of points that satisfy $$f(x,y,z)=3$$ and its gradient will be normal to the sphere.

This gives $$(2x-2)\hat i +(2y-4)\hat j+(2z+2)\hat k=c(2\hat i-2\hat j+\hat k)$$ where $$c$$ is some constant. This gives three equations: $$2x-2=2c\implies x=1+c\\2y-4=-2c\implies y=2-c\\2z+2=c\implies z=\frac{c-2}{2}$$

We can plug these values back into the equation of the plane to get $$c$$.

$$2(1+c)-2(2-c)+(\frac{c-2}{2})+12=0\implies 4c+\frac{c-2}{2}+10=0\implies c=-2$$

Plugging this into our equations for $$x$$, $$y$$ and $$z$$ gives $$(x,y,z)=(-1,4,-2)$$

• $u,v,w$ are not unknowns, but constants to allow for translation from the origin. (They are perhaps poorly named; $A,B,C$ might be better.) Feb 6 at 19:46
• @Théophile Thank you for clarifying that but I probably still wouldn't have understood the approach. I thought he might have been trying to plug the equation of the plane into the equation of the sphere. Feb 6 at 20:04
• @JohnDouma Thank you for your answer! Actually that was a typo. I mistyped it as $(-3,4,0)$ instead of $(3,0,0)$. Even then, $(3,0,0)$ doesn't actually lie on the tangent plane. Thus, by this approach we are not getting the tangent point.I wanted to know, why we are not getting the tangent point by using this approach? Where is the problem occuring? Feb 7 at 4:49

If you, as Alex K suggests above, allow for a scaling factor $$k,$$ $$(xx_1+yy_1+zz_1-(x+x_1)-2(y+y_1)+(z+z_1)-3)-(2x-2y+z+12)k=0$$ you get four equations \begin{align} x_1-2k-1&=0\\y_1+2k-2&=0\\z_1-k+1&=0\\-x_1-2y_1+z_1-12k-3&=0\end{align}

with solution $$k=-1, x_1 =-1 y_1 = 4,z_1 =- 2.$$

• Thank you for your answer! But where is the problem occuring in this approach ? I mean why this approach isn't valid? Feb 7 at 4:51
• @Franklin Because you only set things up to make sure the three variable coefficients in the plane equations match. There is also the constant term in each plane equation, and your setup does not consider how those may not match. So you haven't actually made the one plane equation describe the same plane as the other. (You've merely found one plane that is parallel to the other one.) Feb 7 at 5:16
• @2'59'2 Yes! So the real flaw is, that when I compared the coefficients, I ignored the condtant term in both the equations, and they are not matching in this case atleast. One thing that still bothers me, is that why is it happening so? Feb 7 at 5:19

If you know that $$\langle A,B,C\rangle$$ is a normal vector to the plane

$$A(x-x_1)+B(y-y_1)+C(z-z_1)=0$$

then you may solve the exercise as follows.

Completing the square shows that the center of the sphere is $$(1,2,-1)$$ and the radius is $$r=3$$.

Since a normal vector (i.e. a vector orthogonal to the plane) is

$$\langle 2,-2,1\rangle$$

the point of tangency will have the form

$$(1,2,-1)+(2,-2,1)t=(1+2t,2-2t,-1+t)$$

Substituting into the equation of the plane gives

$$2(1+2t)-2(2-2t)+(-1+t+12=0$$

Solving for $$t$$ gives $$t=-1$$.

Therefore

$$(1+2t,2-2t,-1+t)=(-1,4,-2)$$

NOTE: In general, for a plane which does not cut a sphere, this approach gives either the tangent point or the point of the plane nearest the sphere. In this case you are given that the point is on the sphere, and that is easily checked to be the case.