    Question

# 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 2√3 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 6√2 Given C1:x2+y2=3 intersects the parabola x2=2y On solving x2+y2=3 and x2=2y, we get y2+2y=3 ⇒ y2+2y−3=0 ⇒ (y+3)(y−1)=0 ∴ y=1,−3 [neglecting y=−3,as−√3≤y≤√3] ∴ y=1⇒x=±√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 2√3. ∴ |0+k−3|√2+1=2√3 (Distance of centre Q2 from tangent line)⇒ |k−3|=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=2√3 ∴ R2R3=√d2−(r1+r2)2=√122−(4√3)2=√(144−48)=√96=4√6 ∴ 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=12×4√6×√3=6√2 ∴ Option (c) is correct. Also are ΔPQ2Q3=12Q2Q3×√2=√22×12=6√2 ∴ Option (d) is incorrect  Suggest Corrections  0      Explore more