# How can I find the radius of this circle?

A, B, C, D are points on a circle of centre O and TA is the tangent to the circle at A.

TA = 7.1 cm
AD = 4.5 cm

Basically if we were to draw this circle we would first draw (Clockwise) the point D on the circle then A then C and then B.
Point T being outside of the circle slightly on the right of it.

Now angle BDA = 36 degrees and DAT = 28 degrees

From this info I need to find the radius of the circle and say if OT bisects AD?

I have demonstrated (using the circle property) that angle BOA = 72 degrees and BCA = 144 degrees and also OAD = 62 degrees.

But how can I find the radius and prove the bisection? Please any hints or answers to this treaky one :-)

## Answers and Replies

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Divide AD in two with a point E. Now from the triangle OAE you get OA=AE/cos(62).

Draw the bisector of AD from O to the line AT, call the point it makes on line AT point F. The projection of AF onto AD is AE, from this we get AF=AE/cos(28)=2,55 cm, which is not equal to TA and thus OT does not bisect AD (OF does).

edit: seems deceptively simple, maybe i missed something???

- Kamataat

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seems right thanks

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