Problem: In the adjoining figure TP and Tβ²Q are parallel tangents to a circle of radius r, with T and Tβ² the points of tangency. PTβ²β²Q is a third tangent with Tβ²β² as point of tangency. If TP=4 and Tβ²Q=9 then r is
Answer Choices:
A. 25/6
B. 6
C. 25/4
D. a number other than 25/6,6,25/4
E. not determinable from the given information
Solution:
Since TP=Tβ²P,OT=OTβ²=r, and β PTβ²O=. PTO=90β, we have β³OTPββ³OTβ²P. Similarly β³OTβ²Qββ³OTβ²Q. Letting x=β TOP=β POTβ². and y=Tβ²QQ=βQOTβ² we obtain 2x+2yβ180β But this implies that β POQ=x+y=90β. Therefore β³POQ is a right triangle with altitude OT. Since the altitude drawn to the nypotenuse of a right triangle is the mean proportion of the segments it cuts, we have r4ββ9rβ or r=6.