Lagrange multipliers from hell I was asked to solve this question, decided to try and solve it with lagrange multipliers as I see no other way:
"Find the closest and furthest points on the circle made from the intersection of the ball $(x-1)^2+(y-2)^2+(z-3)^2=9$ and the plane $x-2z=0$ from the point $(0,0)$".
What I did:
the distance for any point $(x,y,z)$ from the origin is $d(x,y,z)=\sqrt{x^2+y^2+z^2}$. so using lagrange multipliers we have:
$d(x,y,z)=\sqrt{x^2+y^2+z^2}$
$C_1(x,y,z)=(x-1)^2+(y-2)^2+(z-3)^2-9$
$C_2(x,y,z)=x-2z$
$L(x,y,z) = d-\lambda_1C_1-\lambda_2C_2 $ meaning:
$L(x,y,z)=\sqrt{x^2+y^2+z^2}-\lambda_1[(x-1)^2+(y-2)^2+(z-3)^2-9]-\lambda_2(x-2z)$
Let's derive and solve when derivatives are zero:
$\frac{\partial L}{\partial x}= \frac{x}{\sqrt{x^2+y^2+z^2}}-2\lambda_1(x-1)-\lambda_2=0$
$\frac{\partial L}{\partial y} = \frac{y}{\sqrt{x^2+y^2+z^2}}-2\lambda_1(y-2)=0$
$\frac{\partial L}{\partial z} = \frac{z}{\sqrt{x^2+y^2+z^2}}-2\lambda_1(z-3)+2\lambda_2=0$
$\frac{\partial L}{\partial \lambda_1} = -(x-1)^2-(y-2)^2-(z-3)^2+9=0$
$\frac{\partial L}{\partial \lambda_2} = 2z-x=0$
Solving this monstrous system seems very unlikely, and very difficult, and not how the question is meant to be solved. am I missing something?

The circle can be parametrized as
$$  x = 2 + \frac{4}{\sqrt 5} \sin \theta, \; y = 2 + 2 \cos \theta, \; z =  1 + \frac{2}{\sqrt 5} \sin \theta.  $$
The squared distance of such a point from the origin is 
$$ f(\theta) = 13 + 4 \sqrt 5 \sin \theta + 8 \cos \theta.  $$
Derivative is
$$ f'(\theta ) = 4 \sqrt 5 \cos \theta - 8 \sin \theta. $$
So, the two extrema occur where $\tan \theta = \frac{\sqrt 5}{2};$ from $1 + \tan^2 \theta = \sec^2 \theta$ we get $$\sec^2 \theta = \frac{9}{4}, \cos^2 \theta = \frac{4}{9},$$ and
$$ \cos \theta = \pm \frac{2}{3}, \; \sin \theta = \pm \frac{\sqrt 5}{3}, $$ with matching $\pm.$
The farthest point from the origin is
$$ \left( \frac{10}{3}, \; \frac{10}{3}, \; \frac{5}{3}     \right),  $$
the nearest
$$ \left( \frac{2}{3}, \; \frac{2}{3}, \; \frac{1}{3}     \right).  $$
The center of the circle is at $(2,2,1),$ and the plane goes through the origin, so all we really did was find the two points on the given sphere along that line, $$ x = 2t,y=2t,z=t. $$
