# 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:

$r=\sqrt{ (40-0)^2+(30-0)^2 }=50 \ \\ x_{0}=|-14|=14 \ \\ \ \\ (x/2)^2 +x_{0}^2=r^2 \ \\ \ \\ x=2 \cdot \ \sqrt{ r^2-x_{0}^2 }=2 \cdot \ \sqrt{ 50^2-14^2 }=96$

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