Problem 75. In a circle of radius $P$, two mutually perpendicular chords $M N$ and $P Q$ (Fig. 78) are drawn. Find the distance between points $M$ and $P$, if $N Q=m$.
Solution. According to Archimedes' Lemma

$$
a^{2}+b^{2}+c^{2}+d^{2}=4 R^{2}
$$

But $\quad a^{2}+c^{2}=M P^{2}$ and $\quad b^{2}+d^{2}=m^{2}$. Then we have: $M P^{2}+m^{2}=4 R^{2}$, from which

![](https://cdn.mathpix.com/cropped/2024_05_21_c833ac06d3d78911cbb1g-062.jpg?height=560&width=508&top_left_y=1008&top_left_x=1185)

$$
M P=\sqrt{4 R^{2}-m^{2}}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_c833ac06d3d78911cbb1g-062.jpg?height=614&width=551&top_left_y=1686&top_left_x=364)