JEE Main Math Class 11 Coordinate Geometry Part 3 Questions

Given, OA=1 unit, OB=13 unit Since, OB is diameter of circle. Then, radius (r)=‌

13

2

=6.5 unitDraw a line joining points P and C, where C is the centre of the given circle.
Then, PC= radius of circle =6.5 units OC= radius of circle =6.5 units
Now, AC=OC−OA=6.5−1=5.5 units Then, using Pythagoras theorem,
(PA)2‌‌=(PC)2−(AC)2
‌‌=(6.5)2−(5.5)2
‌‌=(6.5−5.5)(6.5+5.5)
‌‌=(1)(12)=12
∴‌‌PA‌‌=√12
Then, PQ=2PA=2√12
Hence, area of △PQB=‌

1

2

×( Base ) x (Height)
=‌

1

2

×(PQ)×(AB)
=‌

1

2

×(PQ)×(OB−OA)
=‌

1

2

×(2√12)×(13−1)
=12√12=24√3‌ sq. units ‌