Question

$P$ is a variable point on the line $L=0$. Tangents are drawn to the circle $x^2+y^2=4$ from $P$ to touch it at $Q$ and $R$. The parallelogram $PQRS$ is completed.
If $P\equiv (6,8)$, then the area of $△QRS$ is

• A. $\frac{196\sqrt{5}}{25}$ sq. unit
• B. $\frac{196\sqrt{6}}{52}$ sq. unit
• C. $\frac{192\sqrt{6}}{25}$ sq. unit
• D. $\frac{196\sqrt{5}}{25}$ sq. unit

Answer 1

The correct option is C $\frac{192\sqrt{6}}{25}$ sq. unit


As $PQRS$ is a parallelogram
$\therefore$ area of $△QRS=$ area of $△PQR$
So area of $△QRS=\frac{L^{3}r}{L^2+r^2}$
$=\frac{S_1^{3/2}r}{S_1+r^2}=\frac{(96{)}^{3/2}2}{96+4}=\frac{192\sqrt{6}}{25}$ sq. units

Alternate Solution:
$P\equiv (6,8)$
Equation of $QR$ is $6x+8y=4$
$\Rightarrow 3x+4y-2=0$
$\therefore PM=\frac{48}{5}$ and $PQ=\sqrt{96}$
$QM=\sqrt{96-\frac{(48{)}^2}{25}}=\sqrt{\frac{96}{25}}$
$\therefore QR=2\sqrt{\frac{96}{25}}$
$\therefore \text{area of}△PQR=\frac{1}{2}\cdot PM\cdot QR=\frac{192\sqrt{6}}{25}$
$\because PQRS$ is a rhombus
$\therefore \text{area of}△QRS=\text{area of}△PQR=\frac{192\sqrt{6}}{25}$ sq. units