How do I find the cartesian equation for a circle which is centered around the origin and intersects with the closest single point of a truncus?

Question

The title is poorly worded. Essentially, there is a generic truncus on a cartesian plane, that is $\frac{1}{x^{2}}$.

The question is, how would I find a cartesian equation in order to represent a circle that intersects with the closest possible point on the truncus to the origin. In other words, it has to barely touch the closest edge.

From my limited knowledge of cartesian equations, I know I can use the formula $x^{2}+y^{2}=r^{2}$ in order to get a circle that is on the origin and adjust the value of $r^{2}$ to change the radius of said circle.

How mathematically would I get the $r^{2}$ value (or really whatever that value has to be)? I know the solution is roughly 1.37 but I'd imagine that it's an irrational number.

To help, I've got an image of what I'm looking for: Desmos showing solution

Answer 1

Hint...

Find the equation of the normal to the curve $y=\frac{1}{x^2}$ at the point $(t,\frac{1}{t^2})$. Find the value of $t$ for which this normal passes through the origin.

Can you take it from here?

Answer 2

Well, I got to an answer. I'd say it's not very elegant but it hardly matters. It's just math that happens to work. Variable d is equal to the radius, but by habit I use d for distance.

I found the $d^{2}$ value to be $x^{2}+\frac{1}{x^{4}}$, then taking the derivative of this I ended up with $d^{2}$ prime $= 2x - \frac{4}{x^{5}}$, giving me the x values of $\pm2^{\frac{1}{6}}$. From there it's basic substitution to get $y=\frac{2^{\frac{2}{3}}}{2}$ (which applies to points on both the positive and negative x quadrants).

Then, I took $d=\sqrt{\left(2^{\frac{1}{6}}\right)^{2}+\left(\frac{2^{\frac{2}{3}}}{2}\right)^{2}}$ to give me the distance using pythag, giving me $d=\frac{\sqrt{3}\cdot2^{\frac{2}{3}}}{2}$.

Finally, the cartesian equation for the circle is $x^{2}+y^{2}=\frac{3}{2}\sqrt[3]{2}$.