From P(–4, 0), tangents PA and PA′ are drawn to a circle, x2+y2=4. Where A and A′ is point of contact and A lies above x-axis. Rhombus PAP′A′ is completed.
Column-IColumn-IIColumn-III(I)A≡(−1,√3)(i)PA=2√3(P)P′ lies on the circle, x2+y2=4(II)A′≡(−1,−√3)(ii)Area of ΔPAA′=3√3 sq.units(Q)ΔPAA′ is equilateral(III)P′=(4,0)(iii)PP′=6(R)P′ lies outside the circle , x2+y2=4(IV)P′=(2,0)(iv)Area of ΔPAA′=4√3 sq.units(S)P′ lies inside circle , x2+y2=4
Which one of the following is correct?
(II) (ii) Q
△PAA′ is an equilateral triangle with side length =2√3 square units.
So, area of the triangle PAA′=√34×2√3×2√3=3√3 square units.