Question

Let PQ and RS be tangents at the extremities of the diameter PR of a circle of radius r. PS and RQ intersect at a point X on the circumference of the circle. If PQ = 9 and RS = 4, then the diameter of the circle is :

Answer 1

`/_PRX=theta`
`tantheta=(PQ)/(PR)`
`PR=9Cottheta-(1)`
`tan(90-theta)=(RS)/(PR)`
`PR=4/Cottheta=4tantheta`
`9/4=tan^2theta`
`tantheta=3/2`
`PR=4tantheta=12/2=6`.