UPSC 2017 CDS II Math Paper

Let the radius of the circle be r cm.Remember: The perpendicular drawn from the centre of the circle to the chord bisects the chord.
Now, OE ⊥ AB
⇒AE=BE
=

AB

2

=2cm
Also, OF ⊥ CD
⇒ CF = DF
=

CD

2

=5cm
EF = 3cm
In right ΔOCF,
OF2=OC2−CF2=r2−25....(1)
In right ΔOAE,
OE2=OA2−AE2=r2−4....(2)
Subtracting (1) from (2), we have
(OF+3)2−OF2=21
⇒6OF=21−9=12
⇒OF=2cm
Putting OF = 2 in (1), we have
r2=4+25=29
⇒r=√29cm