$CumbersomeCoordinates-1$

Good one Karthik Sharma, you have used the property that concyclic points lie in a circle.I have used a similar property that the sum of the opposite angles in a concyclic quadrilateral sums to 180 degrees.By formulating the equations of the lines of the quadrilateral, and finding the angle between any two lines and adding it to the angle between the other two lines we would get an equation in terms of A,B.By solvig that equation by hit and trial method we find that (1,-4) is the required point.

As we can see by the given coordinates that the circle intersects the y-axis. Therefore, we can use the equation for the circle intersecting y-axis i.e.

${x}^{2} +{y}^{2} - 2 hx - 2 ky + {h}^{2} =0$

where (k,y) is the coordinate of the centre of the circle.

Using 1st given solution,

13 - 6h + 4k + ${h}^{2}$ =0 -------- 1

Using 2nd given solution,

${(h-1)}^{2} = 0$

h = 1

Putting this value of h in 1,

13 - 6 + 4k + 1 = 0

k = -2

Now substituting value of h and k in the original equation,

${x}^{2} +{y}^{2} - 2x + 4y + 1 = 0$

Now just use hit and trial for all the given choices, we get the answer as $\boxed{(1,-4)}$