Thank you for reporting, we will resolve it shortly
Q.
The point $P \,(3, 6)$ is first reflected on the line $y = x$ and then the image point $Q$ is again reflected on the line $y = - x$ to get the image point $Q'$. Then the circumcentre of the $\Delta PQQ'$ is
The coordinates of $Q$ and $Q^{\prime}$ will be $(6,3)$ and $(-3,-6)$, respectively.
Now, Slope of $P Q=\frac{6-3}{3-6}=-1$
and Slope of $Q Q^{\prime}=\frac{-6-3}{-3-6}=1$
$\therefore $ Slope of $P Q \times$ Slope of $Q Q^{\prime}=-1 \times 1=-1$
$\therefore \triangle P Q Q^{\prime}$ is right angled triangle at $Q$.
$\therefore $ Circumcentre will be the mid-point of hypotenuse
$P Q^{\prime}=\left(\frac{-3+3}{2}, \frac{-6+6}{2}\right)$
$=(0,0)$
