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Two equal circles having centres at $$A$$ and $$B$$ are intersecting in two points $$P$$ and $$Q$$. If $$AB = 8$$ and $$PQ = 6$$ then the radius of the circles is
For two equal intersecting circles, the common chord $$PQ$$ and the line of centres $$AB$$ bisect each other at right angles at a point $$M$$. So $$AM = 4$$ and $$PM = 3$$, and in the right triangle $$AMP$$ the radius is $$AP = \sqrt{4^2 + 3^2} = 5$$.
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