# Is there a constant $C$ such that $z=x^2+y^2+C$ is tangent to $x^2+y^2=z^2$?

Per the title, does a constant $C$ exist such that the surface of the paraboloid $z=x^2+y^2+C$ is tangent to the surface of the cone $x^2+y^2=z^2$? How would I find this constant?

Thanks a lot!

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Since the cone has constant slope, the paraboloid has all slopes in $[0,\infty)$ and you can use $C$ to make them match, the answer to the first question is yes. To answer the second question, it would make things easier to replace $x^2+y^2$ by $r^2$, since they don't appear separately. – joriki May 3 '12 at 17:04
Thanks! That's a useful tip. I'm getting $C=1/4$ for the constant such that the paraboloid is tangnet to the cone, is this accurate? – ro44 May 3 '12 at 17:11
@ro44 Yes, I believe $1/4$ is the answer. – Shuhao Cao May 3 '12 at 17:12
@Jon: Thanks. Also, you shouldn't have deleted your answer, I think, I didn't read it entirely yet but different approaches are always good, right? – ro44 May 3 '12 at 17:14
@ro44 No problem, and my answer is essentially the same as what joriki suggested you to do, since those two surfaces are rotational-invariant with respect to the $z$-axis, meaning if you try to look at them from any perspective perpendicular to the $z$-axis, you can't tell the difference, hence draw a cross-section viewing from $y$-axis of both surfaces, so that you could visualize it as to find $C$ such that at somewhere $z = x^2 + C$ is tangent to $z = \pm x$, the rest is the same. – Shuhao Cao May 3 '12 at 17:22

Yes. $C=1/4$. You can see this by following up on @Joriki's suggestion and solving the equivalent two dimensional problem: $$y=r^2+C=|r|$$ $$y'=2|r|=1$$