The circle C1 :` x^2 + y^2 = 3`, with center at O, intersects the parabola` x^2 = 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 `2sqrt(3)` and centers Q2 and Q3, respectively.If `Q_2 and Q_3` lies on the y-axis, then

Options

A. `Q_(2)Q_(3)=12`
B. `R_(2)R_(3)=4sqrt(6)`
C. area of the triangle `OR_(2)R_(3)" is "6sqrt(2)`
D. area of the triangle `PQ_(2)Q_(3)" is "4sqrt(2)`

Answer

A, B, C

Explanation

1,2,3
`x^(2)=y^(2)=3andx^(2)=2y`
Solving above curves, we get point of intersection as `P-=(sqrt(2),1)`.
Equation of tangent to circle at this point `sqrt(2x)+y=3`.
`:." "tantheta=-sqrt(2)`
`tanalpha=tan(theta-90^(@))=-cottheta=(1)/(sqrt(2))`

So, `sinalpha=(1)/(sqrt(3))=(Q_(3)R_(3))/(Q_(3)T)=(2sqrt(3))/(Q_(3)T)`
`rArrQ_(3)T=6`
`:." "Q_(2)Q_(3)=2Q_(3)T=12` (As circle have equal radii)
`tanalpha=(1)/(sqrt(2))=(Q_(3)R_(3))/(R_(3)T)=(2sqrt(3))/(R_(3)T)`
`rArrR_(3)T=2sqrt(6)`
`rArrR_(2)R_(3)=2R_(3)T=4sqrt(6)`
Perpendiculau distance of O from `R_(2)R_(3)=|(3)/(sqrt((sqrt(2))^(2)+1^(2)))|=sqrt(3)`
`:." Area"(OR_(2)R_(3))=(1)/(2)xxsqrt(3)xx4sqrt(6)=6sqrt(2)` sq. units
Similarly, Area `(PQ_(2)Q_(3))=(1)/(2)xxsqrt(2)xx12=6sqrt(2)` sq. units