Bulgarian National Mathematical Olympiad

Let $O$ be the center of $k$ and $\omega$ be the circle with diameter $AO$. Let $A{A}_1$ and $B{B}_1$ be the altitudes of $△ABC$ and $H$ is the orthocenter. Then the power of $H$ with respect to $\omega$ is equal to $AH\cdot H{A}_1$ and the power of $H$ with respect to $k$ is equal to $BH\cdot H{B}_1$. Since $AH\cdot H{A}_1=BH\cdot H{B}_1$ it follows that $H$ lies on the radical axis $l$ of $k$ and $\omega$. Conversely, it is easy to see that each point of $l$ is an orthocenter of some triangle from given type.

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Alternative solution.

Let $AB\cap k=B^{\prime }$, $AC\cap k=C^{\prime }$ and $B{C}^{\prime }\cap B^{\prime }C=H$. Then $H$ is the orthocenter of $△ABC$ and $H$ lies on the polar $l$ of point $A$ with respect to $k$ (Why?). Conversely, it is easy to see that each point of $l$ is an orthocenter of some triangle from given type.