Concentric Circles

If we draw a radius in the small circle to the point of tangency, it will be at right angle with the chord.(see figure below). If x is half the length of AB, r is the radius of the small circle and R the radius of the large circle then by Pythagora's theorem we have:

r2 + x2 = R2

62 + x2 = 102

Solve for x: x = 8

Length of AB = 2x = 16

If we were to draw a line the length from B to the center point (which is 10) and a line that is the length of the smaller radius to the tangent of the cord AB, we can observe a triangle. All that is left is to use 2(a²+ b² = c² ).

Draw lines from those point of contacts of chord i.e, small circle and big circle to the centre 'O'. B on Big circle and S on small circle. Then you observe a right angled triangle. So from Pythagorean theorem.

OB^2=OS^2+SB^2

SB^2=OB^2-OS^2 = 100-36=64

=> SB = 8

2×SB = 16 = Chord Length .