Question

From $P(–4,0)$, tangents $PA$ and $P{A}^{\prime }$ are drawn to a circle, $x^2+y^2=4$. Where $A$ and $A^{\prime }$ is point of contact and $A$ lies above x-axis. Rhombus $PA{P}^{\prime }{A}^{\prime }$ is completed.

$\begin{array}{cccccc}& \text{Column-I}& & \text{Column-II}& & \text{Column-III}\\ \left(I\right)& A\equiv (-1,\sqrt{3})& \left(i\right)& PA=2\sqrt{3}& \left(P\right)& P^{\prime }\text{lies on the circle},x^2+y^2=4\\ \left(II\right)& A^{\prime }\equiv (-1,-\sqrt{3})& \left(ii\right)& \text{Area of}\Delta PA{A}^{\prime }=3\sqrt{3}\text{sq.units}& \left(Q\right)& \Delta PA{A}^{\prime }\text{is equilateral}\\ \left(III\right)& P^{\prime }=(4,0)& \left(iii\right)& P{P}^{\prime }=6& \left(R\right)& P^{\prime }\text{lies outside the circle},x^2+y^2=4\\ \left(IV\right)& P^{\prime }=(2,0)& \left(iv\right)& \text{Area of}\Delta PA{A}^{\prime }=4\sqrt{3}\text{sq.units}& \left(S\right)& P^{\prime }\text{lies inside circle},x^2+y^2=4\end{array}$

Which one of the following is correct?

• A. (I) (iv) Q
• B. (I) (ii) R
• C. (IV) (iv) P
• D. (IV) (iii) P

Answer 1

The correct option is D

(IV) (iii) P

$P{P}^{\prime }=6$ and $P\equiv (-4,0)$

So, $P^{\prime }\equiv (2,0)$

Putting the point, $P^{\prime }(2,0)$ in the equation of circle, $x^2+y^2-4=0$, we get,

$2^2+0-4=0$

Hence, $P^{\prime }$ lies on the circle

The correct combination is (IV) (iii) P