Solution:
According to the given question,
Let \[O\text{ }and\text{ }O\]be the centres of two intersecting circles, where points of the intersection are \[P\text{ }and\text{ }Q\text{ }and\text{ }PA\text{ }and\text{ }PB\] are their diameters respectively.
Join \[PQ,\text{ }AQ\text{ }and\text{ }QB\]
Hence, \[\angle AQP\text{ }=\text{ }{{90}^{o}}~and\angle BQP\text{ }=\text{ }{{90}^{o}}\]
[Angle in a semicircle is a right angle]
Adding both these angles we get
\[\angle AQP\text{ }+\angle BQP\text{ }=\text{ }{{180}^{o}}\]
\[\angle AQB\text{ }=\text{ }{{180}^{o}}\]
Therefore, the points \[A,\text{ }Q\text{ }and\text{ }B\]are collinear.