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Let x^2 +y^2 -2x-2y-2=0 and x^2 +y^2 -6x...

Let `x^2 +y^2 -2x-2y-2=0` and `x^2 +y^2 -6x-6y+14=0` are two circles `C_1, C_2` are the centre of circles and circles intersect at `P,Q` find the area of quadrilateral `C_1 P C_2 Q`

A

8

B

4

C

6

D

9

Text Solution

Verified by Experts

The correct Answer is:
B
Given circles,
`x^(2)+y^(2)-2x-2y-2=0 " "...(i)`
and `x^(2)+y^(2)-6x-6y+14=0" "...(ii)`
are intersecting each other orthogonally, because
2(1)(3)+2(1)(3)=14-2
`[therefore " two circle are intersected orthogonally if "2g_(1)g_(2)+2f_(1)f_(2)=c_(1)+c_(2)]`

So, area of quadrilateral
`PC_(1)QC_(2)=2xxar(DeltaPC_(1)C_(2))`.
`=2xx((1)/(2)xx2xx2)=4` sq units.
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