if the parametric equations of the circle passing through the points $(3,4), (3, 2)$ and 

$(1,4)\text{ is }x=a+r\cos \theta ,y=b+r\sin \theta \text{, then }{b}^a{r}^a=$

$(3, 4), (3,2)$ lie on vertical line.

$(3,4),(1,4)$ lie on horizontal line. 

$\therefore$  Given points form a right angled triangle with vertex at $(3, 4)$.

$(3, 2), (1, 4)$ are the ends of  the diameter of the circle passes through the given points. 

The equation of the circle is $(x-3)(x-1)+(y-2)(y-4)=0\Rightarrow x^2+y^2-4x-6y+11=0$

Centre $=(2,3),\text{ radius }=\sqrt{4+9-11}=\sqrt{2}$

Parametric equations are $(2+\sqrt{2}\cos \theta ,3+\sqrt{2}\sin \theta ).\therefore a=2,b=3,r=\sqrt{2}$

Now $b^a{r}^a=3^2(\sqrt{2}{)}^2=9\times 2=18$.