Two equal circles intersect each other in such a way that the centre of each circle lies on the circumference of the other. If the length of their common chord is 10√3 cm, what is the diameter of each circle in centimetres?

Introduction / Context:
This geometry question deals with two equal circles that intersect in a special way: the centre of each circle lies on the circumference of the other. This configuration is sometimes called a symmetric or lens shaped intersection. The problem provides the length of the common chord of the two circles and asks for the diameter of each circle. Solving it requires knowledge of basic circle geometry and right triangle relationships.

Given Data / Assumptions:

• The two circles are equal, so both have the same radius r.
• Each circle passes through the centre of the other, so the distance between the centres is equal to r.
• The length of the common chord of intersection is 10√3 cm.
• We assume the usual Euclidean geometry properties of circles.
• We need to find the diameter, which is 2r.

Concept / Approach:
When two equal circles intersect, the line segment joining their centres is the perpendicular bisector of their common chord. In this special configuration, the distance between the centres is equal to the radius r. The common chord lies at equal distance from both centres. We can draw a right triangle whose hypotenuse is r (from centre to any end of the chord) and whose one leg is half of the chord length. Using the Pythagorean theorem, we solve for r.

Step-by-Step Solution:
Step 1: Let r be the radius of each circle.Step 2: The distance between the two centres is given to be equal to r, because each centre lies on the circumference of the other circle.Step 3: Let P and Q be the points of intersection of the two circles, and let AB be the common chord with endpoints A and B.Step 4: The line joining the centres is perpendicular to AB and bisects it.Step 5: The length of the chord AB is 10√3 cm, so half the chord is AB/2 = 5√3 cm.Step 6: Consider the right triangle formed by one centre, the midpoint of AB, and one endpoint of AB. The hypotenuse is r and one leg is half the chord, 5√3.Step 7: By the Pythagorean theorem, r^2 = (5√3)^2 + d^2, where d is the distance from the centre to the chord. But in this special configuration the distance between centres is r, and geometric symmetry shows that d = r/2.Step 8: Another direct approach is to use the standard relation for an equal intersecting circle configuration: chord length = r√3, so r√3 = 10√3.Step 9: Cancel √3 on both sides to get r = 10 cm.Step 10: Therefore the diameter is 2r = 20 cm.

Verification / Alternative check:
To verify, note that if r = 10, then the distance between centres is also 10. In a right triangle formed by the centre of one circle, the midpoint of the chord, and an endpoint of the chord, we have hypotenuse 10 and one leg 5√3. The remaining leg equals √(10^2 − (5√3)^2) = √(100 − 75) = √25 = 5. That distance represents how far the chord is from the centre, which is consistent in both circles. This confirms that r = 10 is geometrically valid and that the chord length 10√3 is correctly matched.

Why Other Options Are Wrong:
A diameter of 10 cm would correspond to a radius of 5 cm, which would make the chord shorter than 10√3 cm. Larger values like 30 cm or 25 cm give radii that produce significantly longer chords than the one specified. The 15 cm diameter also does not yield a chord of 10√3 cm in this configuration. Only a radius of 10 cm, resulting in a diameter of 20 cm, satisfies the geometric constraints of the intersecting circles and the given chord length.

Common Pitfalls:
A common mistake is to treat the distance between the centres as 2r instead of r when each centre lies on the circumference of the other circle. Another error is to forget that the chord is bisected by the line joining the centres and therefore half of the chord should be used in right triangle calculations. Careful interpretation of the configuration and correct use of Pythagoras prevent these errors.

Final Answer:
The diameter of each circle is 20 cm.