1. **Label the given conic as $\mathcal{H}$ and identify the tangent line $\tau$ to $\mathcal{H}$ at point $O$.**
- Let $\omega$ be an arbitrary fixed circle tangent to $\tau$ at $O$.
- Point $O$ is the center of a homology $\mathcal{U}: \omega \mapsto \mathcal{H}$.

2. **Consider the points $A$ and $B$ where the right angle with vertex $O$ intersects the conic $\mathcal{H}$.**
- Let $OA$ and $OB$ cut $\omega$ again at $A'$ and $B'$, respectively. These points $A'$ and $B'$ are the homologous points of $A$ and $B$ under the homology $\mathcal{U}$.

3. **Analyze the angles formed by the points $A'$, $B'$, and $O$.**
- Since $\angle AOB = 90^\circ$, it follows that $\angle A'OB' = 90^\circ$ as well.

4. **Determine the line $A'B'$ in relation to the fixed point $P$.**
- Because $\angle A'OB' = 90^\circ$, the line $A'B'$ always passes through the fixed center $P$ of the circle $\omega$.

5. **Map the fixed point $P$ under the homology $\mathcal{U}$.**
- The image of the fixed point $P$ under the homology $\mathcal{U}$ is denoted as $P'$.

6. **Conclude the behavior of the line $AB$.**
- Since $A'B'$ always passes through $P$, the line $AB$ (the image of $A'B'$ under $\mathcal{U}$) always passes through the fixed point $P'$.

7. **Relate the fixed point $P'$ to the normal line to $\mathcal{H}$ through $O$.**
- Consequently, the chords $AB$ pass through a fixed point on the normal line to $\mathcal{H}$ through $O$.

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