Chord BC

A circle k has the center at the point S = [0; 0]. Point A = [40; 30] lies on the circle k. How long is the chord BC if the center P of this chord has the coordinates: [- 14; 0]?

Correct result:

x =  96

Solution:

$$\begin{gathered} r=\sqrt{(40-0{)}^2+(30-0{)}^2}=50 \\ {x}_0=\mathrm{\mid }-14\mathrm{\mid }=14 \\ (x\mathrm{/}2{)}^2+x_0^2=r^2 \\ x=2\cdot \sqrt{r^2-x_0^2}=2\cdot \sqrt{50^2-14^2}=96 \end{gathered}$$

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