Given that the circles intersect at two points, so we can draw the above figure. Let AB be the common chord. Let O and O’ be the centers of the circles, respectively.
O’A = 5 cm, OA = 3 cm
OO’ = 4 cm [Given distance between the centres is 4cm]
Since the radius of the bigger circle is more than the distance between the 2 centers, we can say that the center of the smaller circle lies inside, the bigger circle itself.
OO’ is the perpendicular bisector of AB.
So, OA = OB = 3 cm
AB = 3 cm + 3 cm = 6 cm [Since, O is the mid point of AB]
The length of the common chord is 6 cm.
It is also evident that the common chord is the diameter of the smaller circle.
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Answer:
If two circles of radii 5cm and 3cm intersect at two points and the distance between their centers is 4cm, then the length of the common chord is 6cm.
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Step-by-step explanation:
Given that the circles intersect at two points, so we can draw the above figure. Let AB be the common chord. Let O and O’ be the centers of the circles, respectively.
O’A = 5 cm, OA = 3 cm
OO’ = 4 cm [Given distance between the centres is 4cm]
Since the radius of the bigger circle is more than the distance between the 2 centers, we can say that the center of the smaller circle lies inside, the bigger circle itself.
OO’ is the perpendicular bisector of AB.
So, OA = OB = 3 cm
AB = 3 cm + 3 cm = 6 cm [Since, O is the mid point of AB]
The length of the common chord is 6 cm.
It is also evident that the common chord is the diameter of the smaller circle.