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) (ii) R
• B. (I) (iii) Q
• C. (IV) (ii) (S)
• D. (I) (iii) R

Answer 1

The correct option is B

(I) (iii) Q

The equation of chord of contact is $-4x-4=0$

$\Rightarrow x=-1$

Putting the value of $x$ in the equation of circle, we get,

$1+y^2=4$

$\Rightarrow y=\pm \sqrt{3}$

So, $A\equiv (-1,\sqrt{3})$ and $A^{\prime }\equiv (-1,-\sqrt{3})$

Let the point at which $A{A}^{\prime }$ intersects the x-axis be $Q$

Since, $A{A}^{\prime }$ is perpendicular to x-axis, $Q\equiv (-1,0)\Rightarrow PQ=|-4+1|=3$

$\because PA{P}^{\prime }{A}^{\prime }$ is a rhombus $\therefore P{P}^{\prime }=2PQ$

$\Rightarrow P{P}^{\prime }=6$

Lengths of tangents, $PA=P{A}^{\prime }=\sqrt{(-4{)}^2+0-4}=2\sqrt{3}$

$A{A}^{\prime }=2\sqrt{3}$

Hence, $△PA{A}^{\prime }$ is an equilateral triangle.

Thus, correct combination is (I) (iii) Q