Must $P$ be a circle?

Question

Let $P$ be a convex curve inside a circle $C$, such that for every point $x$ on $C$, there are two perpendicular tangent lines of $P$ go through $x$. Must be $P$ be a circle?

If we change the condition "two tangent lines are perpendicular" by "the angle between them is $\frac{\pi}{3}$", the problem become easy: Let $x$ be a point on $C$, let two tangent lines of $P$ which go through $x$ intersect $C$ again at $y,z$, then the two tangent lines of $P$ which go through $y$ and $z$ respectively but not $x$ are either parallel, which can't be because a convex curve $P$ can't have two parallel tagent lines such that $P$ is not between them, or they are coincide, which means $xyz$ is an equilateral triangle. So for every equilateral triangle which vertices on $C$, all of their edges are tangent to $P$ and it is easy to see that $P$ must be a circle.

But the problem is not simple if the angle between them is $\frac{\pi}{2}$ or more generally $\frac{(2m-1)\pi}{2n},m,n\in\mathbb{N}^*$. The reason is that if you try to find two tangent lines of $P$ which parallel such that $P$ is not between them by above way, they are actually always coincide.

Answer 1

For the case where the tangents are perpendicular the inner curve may be an ellipse. To wit:

1. Inscribe a square in the circle

2. Define the center of the ellipse to match that of the circle and square and the foci to both lie on one diagonal of the square.

3. Set the major axis of the ellipse (sum of distances from the two foci) so that it's tangent to one side of the square, in which case by symmetry it will be tangent to all four sides of the square.

4. We have defined the tangents to be perpendicular from just four points on the circle, but in fact the tangents from all points on the circle to the ellipse will be perpendicular. See this Wikipedia article.

We may parametrize the ellipse in the following manner ($r$ is the radius of the outer circle, $\theta$ is any constant between $0$ and $\pi/4$):

Semimajor axis = $r\cos\theta$

Semiminor axis = $r\sin\theta$

Eccentrivity = $\sqrt{\cos(2\theta)}/\cos\theta$

The reader may want to identify the limiting construction where $\theta\to0$.

Answer 2

As an addendum to Oscar's answer:

The authors of

Cieślak, Waldemar; Mozgawa, Witold, On curves with circles as their isoptics, Aequationes Math. 96, No. 3, 653-667 (2022). ZBL1497.53005.

gave a complete characterization (in terms of Fourier series of the associated periodic functions $p(t)$) of $C^1$-smooth closed convex planar curves $P$ for which the isoptic curve is a round circle (your circle $C$). In particular, there are non-conic curves $P$ with this property. As a specific example, they give the curve $P$ in the complex plane with the parameterization $$ z(t)= p(t) e^{it} + i p'(t)e^{it}, t\in [0, 2\pi], $$ where $$ p(t)= \sqrt{a \sin^2(3t) + b \cos^2(9t) +c,} $$ for fixed values of the parameters $a, b, c$.