Problem
GEO-B2-M09-P022 Chord Length from Distance to Centre
#22
★★★★☆ Level 4 of 5
The circle \(x^2+y^2=25\) intersects the line \(3x+4y=20\) at points \(A\) and \(B\). Find the midpoint of \(AB\) and the length \(AB\).
The midpoint of a chord is the foot of the perpendicular from the centre of the circle to the line.
The centre is \(O(0,0)\), and the radius is \(5\). The foot of the perpendicular from \(O\) to the line \(3x+4y=20\) is \(M\left(\frac{3\cdot 20}{3^2+4^2},\frac{4\cdot 20}{3^2+4^2}\right)=\left(\frac{12}{5},\frac{16}{5}\right)\). The distance \(OM=4\). Half the chord is \(\sqrt{5^2-4^2}=3\), so \(AB=6\).
Nicely shows that coordinates and ordinary geometry work together.