# 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]?

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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Tips to related online calculators
For Basic calculations in analytic geometry is helpful line slope calculator. From coordinates of two points in the plane it calculate slope, normal and parametric line equation(s), slope, directional angle, direction vector, the length of segment, intersections the coordinate axes etc.
Pythagorean theorem is the base for the right triangle calculator.

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