A triangle ABC  with its interior angles ${90}^{°},{15}^{°},{75}^{°}$  is placed such that it is right angled vertex A,  lies on the positive y-axis and its two sides touch the ellipse  $x^2+3y^2=3$, while the hypotenuse BC,  touches the auxiliary circle of that ellipse. Then which of the following is / are true

Note that $P$  lies on the director circle  $x^2+y^2=4$. The equation of P R can be written as $\frac{x}{\sqrt{3}}\cos +y\sin \theta =1$  for some  $\theta$. 
Since the slope of P R is  $-1,\sin \theta =\frac{1}{2}$ and so the equation of P R is $x+y=2$ .
 This implies that the equation of P Q is $y-x=2$ . 
The slope of Q R is $-\frac{1}{\sqrt{3}}$ , while using the fact that Q R is tangent to the auxiliary circle $x^2+y^2=3,$ its equation can be written as $x\cos {\theta }_1+y\sin {\theta }_1=\sqrt{3}$  for some ${\theta }_1$ . 
Thus, $-\cot {\theta }_1=-\frac{1}{\sqrt{3}}$ or ${\theta }_1={60}^{°}$ , and so the equation of Q R is $x+\sqrt{3}y=-2\sqrt{3}$ .
We now have the equations of the three sides of  $△PQR$, from which we can obtain all the points:
$P\equiv (0,2),Q\equiv (-6+2\sqrt{3},-4+2\sqrt{3}),R\equiv (6+2\sqrt{3},-4-2\sqrt{3})$
The centroid is $(\frac{4}{\sqrt{3}},-2)$  while the circumcenter is the mid-point of Q R, i.e.,  $(2\sqrt{3},-4)$.