Q17 of 31 Page 65

Two equal circles of radius 10 cm. intersect each other and the length of their common chord is 12 cm. Let us determine the distance between the two centers of two circle


Let Center of the Circles are A and B. CD is a common Chord of the circle. AB is the perpendicular bisector of the chord CD.


If AB is a perpendicular bisector of CD then it should passes through both the centers.


So, AB is the distance that we need to calculate.


Given, AC = 10cm, CD = 12cm


⇒ CM


⇒ CM


⇒ CM = 6cm


In ACM


⇒ AC2 = AM2 + CM2


⇒ AM2 = AC2-CM2


⇒ AM2 = 100-36


⇒ AM2 = 64


⇒ AM = 8cm


In BCM


⇒ BC2 = BM2 + CM2


⇒ BM2 = BC2-CM2


⇒ BM2 = 100-36


BM2 = 64


BM = 8cm


AB = 16cm


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16

Let us write True/False:

i. Only one circle can be drawn through three collinear points


ii. The two circles ABCDA and ABCEA are same circle.


iii. If two chord AB and AC of a circle with its centre O are situated on the Opposite side of the radius OA, then ∠OAB=∠OAC.

 16

Let us fill in the blanks:

i. If the ratio of two chords PQ and RS of a circle with its centre O is 1 : 1, then ∠POQ: ∠ROS=_______________


ii. The perpendicular bisector of any chord of a circle is __________ of that circle.

 17

AB and AC are two equal chords of a circle having the radius of 5 cm. The centre of the circle

is situated at the outside of the triangle ABC. If AB = AC = 6 cm. then let us calculate the length of the chord BC.

 17

The lengths of two chords AB and CD of a circle with its centre O are equal. If ∠AOB = 60° and CD = 6 cm. then let us calculate the length of the radius of the circle.