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 $PQSR$ is completed.
On the basis of the above information, answer the following questions:If $P\equiv (2,3)$, then the centre of circumcircle of triangle $QRS$ is

• A. $(\frac{2}{13},\frac{7}{26})$
• B. $(\frac{2}{13},\frac{3}{26})$
• C. $(\frac{3}{13},\frac{9}{26})$
• D. $(\frac{3}{13},\frac{2}{13})$

Answer 1

The correct option is C $(\frac{3}{13},\frac{9}{26})$

$\because$ Length of tangents from a point to circle are equal.

$PQ=PR$

Then parallelogram $PQRS$ is rhombus.

$\therefore$ Mid-point of $QR=$ midpoint of $PS$

and $QR\perp PS$

$\therefore S$ is the mirror image of $P$ w.r.t. $QR$

$\because L\equiv 2x+y=6$

Let $P\equiv (\lambda ,6-2\lambda )$

$\because \angle PQO=\angle PRO=\frac{\pi }{2}$

$\therefore OP$ is diameter of circumcircle $PQR$ then centre is

$(\frac{\lambda }{2},3-\lambda )$

$\therefore x=\frac{\lambda }{2}\Rightarrow \lambda =2x$

and $y=3-\lambda ,$

then $2x+y=3$$\because P\equiv (2,3)$

$\therefore$ Equation of $QR$ is $2x+3y=4$

Let $S\equiv (\alpha ,\beta )$

$\Rightarrow \frac{\alpha -2}{2}=\frac{\beta -3}{3}=\frac{-2(4+9-4)}{(4+9)}=\frac{-18}{13}$

$\therefore \alpha =\frac{10}{-13},\beta =\frac{-15}{13}$

Now, equation of circumcircle of $△QRS$ is
$(x^2+y^2-4)+\lambda (2x+3y-4)=0$

$\because$ it passes through $S\equiv (\frac{-10}{13},\frac{-15}{13})$ then

$(\frac{100}{169}+\frac{225}{269}-4)+\lambda (-\frac{20}{13}-\frac{45}{13}-4)=0$

$\frac{-351}{169}+\lambda (-9)=0$

or $-9\lambda =\frac{-351}{169}$

$\therefore \lambda =-\frac{39}{169}=-\frac{3}{13}$

$\therefore$ Circumcentre is $(-\lambda ,-\frac{3\lambda }{2})\equiv (\frac{3}{13},\frac{9}{26})$