Question

P is a variable point on the line `L=0` . Tangents are drawn to the circles `x^(2)+y^(2)=4` from P to touch it at Q and R. The parallelogram PQSR is completed.
If `P-= (6,8)`, then the area of `Delta QRS` is
A. `(3sqrt(6))/(25)` sq. units
B. `(3sqrt(24))/(25)` sq. units
C. `(48sqrt(6))/(25)` sq. units
D. `(192sqrt(6))/(25)` sq. units

Answer 1

Correct Answer - 4
`P -= (6,8)`
Thereforem the equation of QR (chord of contact ) is
`6x+8y=4`
or `3x+4y-2=0 `
`:. PM =(48)/(50` and `PQ =sqrt(96)`
`QM =sqrt(96-(48^(2))/(25))=sqrt((96)/(25))`
`:. QR = 2 sqrt((96)/(25))`
`:. ` Area of `Delta PQR =(1)/(2) xx PM xx QR =(192 sqrt(6))/(25)`
PQRS is a rhombus. Therefore,
Area of `Delta RS =` Area of `Delta PQR = (192sqrt(6))/(25)`