1. A tangent PQ at a point P of a circle of radius 5cm meets a line through the centre O at a Q so that OQ = 13cm. Find the length PQ.

2.From a point Q, the length of the tangent to a circle is 24cm and the distance of Q from the centre is 25cm.,find the radius of the circle.

3.OD is perpendicular to the chord AB ofthe circle whose center is O.If BC is diameter,find CA/OD=?


given radius OP=5 cm, and OQ = 13 cm

PQ is the tangent to the circle. 

∠OPQ=90 deg [angle joining the center and the point of contact]

in the right angled ΔOPQ, by the pythagoras theorem


(ii) similarly in the 2nd question: PQ=24 cm and OQ=25 cm,

here we need to find the radius OP.


Given : OD is perpendicular to chord AB where O is the center of Circle and BC is diameter.

to find the CA/OD

Construction : join O to A

since ∠CAB=90 deg [angle formed in the semicircle]

∠ODB=90 deg, OD and CA are perpendicular to AB

therefore ,

since O is the mid-point of BC.[OB=OC=radius]

therefore by the mid-point theorem D is the mid-point of AB.

hence ΔBOD and ΔBCA are similar triangles.

therefore ratio of their corresponding sides will be equal.

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 1. pqo forms a right angled triangle as we know that tamgent at the corcle forma a 90 degree angle with radius at the point of contact..

therefore length of pq , we will find by pythagorus theorem 

pq sq. = qo sq. - po sq.

pq sq. = 13 sq. - 5 sq. = 169 - 25 = 144

therefore pq = root of 1444 = 12 ..

therefore length of pq = 12 cm.

2.similarily as in 1. ques... we'lll use pythagorus theorem here using the same reason of forming a right triangle by tangent :

but here radius is not given instead tangent is given and we have to fing radius =

po sq. = oq sq. - qp sq.

po sq. = 25 sq. - 24 sq. = 49

therefore po = root of 49 = 7 ...

therefore length of radius of the circle = 6 cm.

3. i m confused for this part ...

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