Step-by-step Solution:

\[Assertion A:\] "If two circles intersect at two points, then the line joining their centers is perpendicular to the common chord." ✅ True - Consider two intersecting circles with centers \( O_1 \) and \( O_2 \), and let their common chord be \( AB \). - The key fact is that the line joining the centers of the two circles is the perpendicular bisector of the common chord. - This follows from the property that any two intersecting circles form a symmetric configuration where the radical axis (the common chord) is perpendicular to the line joining the centers. --- \[Reason R:\] "The perpendicular bisectors of two chords of a circle intersect at its center." ✅ True - The perpendicular bisector of any chord of a single circle must pass through the center. - Given two different chords, their perpendicular bisectors must intersect at the unique center of the circle. --- \[Does R correctly explain A \] Now, we need to check if Reason R logically supports Assertion A. - Consider two intersecting circles. - The common chord belongs to both circles. - The perpendicular bisector property of chords in a single circle applies to each circle separately. - Since the perpendicular bisectors of chords must pass through the center, the line joining the centers of two circles must be perpendicular to the common chord. - Thus, the perpendicular bisector property (stated in R) leads to the result in A, meaning R correctly explains A. - Final Answer: \[ \boxed{A} \quad (Both A and R are correct, and R correctly explains A.) \] This confirms that option A is indeed correct.