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I've edited the question, now it make sense LOL!!
AO=OB=OE=r
ED=5
DO=5-r

\((5-r)^2+(2.5)^2=r^2\)
r=3.125

(5-r) is positive


chetan2u
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AB is a chord of a circle. AB = 5 cm. A tangent parallel to AB touches the minor arc AB at E. What is the radius of the circle?

1. AB is not a diameter of the circle.
2. The distance between AB and the tangent at E is 5 cm.


Statement 1 does not mean much..

2. The distance between AB and the tangent at E is 5 cm


We can draw a triangle by joining E with the center. This line OE will bisect chord AB at 90'.
The triangle AOG will give a value of r ..
\(r^2=(r-5)^2+2.5^2......r^2=r^2+25-10r+6.25.....10r=31.25....r=3.125\)

Although B is s the answer, but values are not proper as we get r-5 a negative quantity.

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Why is stetment one not needed. If one is true the reaius will be 2.5 . Is it because in the stem minor arc has been mentioned, implying that AB is not a diameter
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Is it right to assume that the tangent is drawn on minor axis? If it was on major axis AB, then the radius would have been different

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Hi chetan2u ,
The question has been modified. Now the stem does not mention intersection on either minor or major arc. I feel it should be C now. Could you please revisit this question ?


Regards

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ShubhamG
Is it right to assume that the tangent is drawn on minor axis? If it was on major axis AB, then the radius would have been different

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Actually not, even if the tangent is on the major arc, then also it would be the same. Please see attached to find the solution -
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Do I actually need to calculate the radius to know that statement 2 is enough? Doesn't fixing three points of a circle (by means of relative distances of point E, A, and B) automatically fix the radius of the circle?

Can anyone let me know if my reasoning is flawed?
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