By theorem, the perpendicular from the center of a circle to a chord bisects the chord. So, C'C bisects AB. Since CC' ⊥ AB and line segment joining the centers of two circles bisects the common chord. Therefore, C' is the mid-point of AB.
By theorem, the perpendicular from the center of a circle to a chord bisects the chord. So, C'C bisects AB. Since CC' ⊥ AB and line segment joining the centers of two circles bisects the common chord. Therefore, C' is the mid-point of AB.
kookie i am fine just my mom don't let me take phone after my exams are over that's why
on Monday my result will also release and please see my questions also I saw that you don't see my questions) :
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Step-by-step explanation:
By theorem, the perpendicular from the center of a circle to a chord bisects the chord. So, C'C bisects AB. Since CC' ⊥ AB and line segment joining the centers of two circles bisects the common chord. Therefore, C' is the mid-point of AB.
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Answer:
By theorem, the perpendicular from the center of a circle to a chord bisects the chord. So, C'C bisects AB. Since CC' ⊥ AB and line segment joining the centers of two circles bisects the common chord. Therefore, C' is the mid-point of AB.
kookie i am fine just my mom don't let me take phone after my exams are over that's why
on Monday my result will also release and please see my questions also I saw that you don't see my questions) :