Question

2 circles of radii x cm and y cm (x greater than y) intersect at 2 pts. P and Q respectively. If the distance 'd' between the centres of the 2 circles is given by [d square =x square + y square], prove that the length of the common chord is 2y cm

Answer 1

See diagram.

d² = x² - y²   and not plus .. correction.

Given C1 P = x,    C2 P = y,    PR = common chord.
C1 C2 = d .     Let the point of intersection of PR with C1 C2  be   Q.

Given  C1C2² = d²  = x² - y²

In the Δ C1C2P, we have
           C1P² = C2P² + C1C2²   because,  x² = y² + (x² - y²) 

 Hence ΔC1C2P is a right angle triangle at C2.   Q and C2 are the same point.  Hence ∠C1C2R and ∠C1C2P are right angles.  Hence, PR is same as the straight line PC2R, which is the diameter of C2.

So the length of PC2R = PC2+ C2R = y + y = 2y.