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The circle C1:x2+y2=3 having centre at origin,O intersects the parabola x2=2y at the point P in the first quadrant. Let the tangent to the circle C1 at P touches other two circles C2 and C3 at R2 and R3, respectively. Suppose C2 and C3 have equal radii 23 and centres Q2 and Q3, respectively. If Q2 and Q3 lie on the Y-axis then


A

Q2Q3=12

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B

R2R3=46

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C

area of the ΔOR2R3 is 62

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D

area of the ΔPQ2Q3 is 42

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Solution

The correct option is C

area of the ΔOR2R3 is 62


Given C1:x2+y2=3 intersects the parabola x2=2y


On solving x2+y2=3 and x2=2y, we get
y2+2y=3
y2+2y3=0
(y+3)(y1)=0
y=1,3 [neglecting y=3,as3y3]
y=1x=±2
P(2,1)ϵ I quadrant
Equation of tangent at P(2,1) to C1:x2+y2=3 is
2x+1.y=3(i)
Now, let the centres of C2 and C3 be Q2 and Q3, and tangent at P touches C2 and C3 at R2 and R3 shown as below



Let Q2 be (0, k) and radius is 23.
|0+k3|2+1=23 (Distance of centre Q2 from tangent line) |k3|=6 k=9,3 Q2(0,9) and Q3(0,3)
Hence, Q2Q3=12
Option (a) is correct.
Also R2R3 is common internal tangent to C2 and C3
And r2=r3=23
R2R3=d2(r1+r2)2=122(43)2=(14448)=96=46


Option (b) is correct
Length of perpendicular from O(0, 0) to R2R3 is equal to radius of C1=3.
Area of ΔOR2R3=12×R2R3×3=12×46×3=62
Option (c) is correct.
Also are ΔPQ2Q3=12Q2Q3×2=22×12=62
Option (d) is incorrect


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