JEE Main 12 Jan 2019 Morning Question 3
Question: Let $ C _1 $ and $ C _2 $ be the centres of the circles $ x^{2}+y^{2}-2x-2y-2=0 $ and $ x^{2}+y^{2}-6x-6y+14=0 $ respectively. If P and Q are the points of intersection of these circles, then the area (in sq. units) of the Quadrilateral $ PC _1QC _2 $ is [JEE Main Online Paper Held On 12-Jan-2019 Morning]
Options:
A) 8
B) 4
C) 6
D) 9
Show Answer
Answer:
Correct Answer: B
Solution:
Let $ S _1\equiv x^{2}+y^{2}-2x-2y-2=0 $ $ S _2\equiv x^{2}+y^{2}-6x-6y+14=0 $
$ \therefore $ $ C _1\equiv (1,1) $ and $ C _2\equiv (3,3). $ Radius of $ S _1,PC _1=\sqrt{1+1+2}=2 $ Radius of $ S _2,PC _2=\sqrt{9+9-14}=2 $
$ \therefore $ $ PC _1=QC _1=PC _2=QC _2=2 $ Now, $ 2g _1g _2+2f _1f _2 $ $ =2\times 3+2\times 3=6+6=12 $ and $ c _1+c _2=14-2=12 $
Here, $ 2g _1g _2+2f _1f _2=c _1+c _2 $
$ \therefore $ Both circles are orthogonal. So, $ PC _1QC _2 $ is a square.
$ \therefore $ Area of $ PC _1QC _2=2\times 2=4sq. $ units
