# O1 AND O2 ARE THE CENTRES OF TWO CONGRUENT CIRCLES INTERSECTING EACH OTHER AT POINTS C AND D. THE LINE JOINING THEIR CENTRES INTERSECTS THE CIRCLES AT A AND B SUCH THAT AB > O1O2. IF CD = 6 CM AND AB = 12 CM FIND THE RADIUS OF THE EITHER CIRCLE.

As circle are congruent , So they both have same radius .  and AB  > O1 O2 , SO we form our diagram , As :

Here
O1C = O1D = O2C = O2C = O1A = O2B  =  r
And
From O1C = O1D = O2C = O2C ,

We get Quadrilateral  O1CO2D is a rhombus .

And we know diagonal of rhombus are perpendicular bisector , So

CD $\perp$O1O2 And CD bisect O1O2 , So

CP  =
And
O1P =
Now triangle O1PC  , we apply Pythagoras theorem , and get

O1C2 = O1P2 + CP2  , Substitute values , we get

r2 = ( 6 - r )2 + 32

r2 = 36 + r2 - 12r  + 9

12r  = 45

r =  3.75
So,
Radius of either circle  =  3.75 cm                                                     ( Ans )

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