Q.Show that the points $(9,1)$, $(7,9)$, $(-2,12)$ and $(6,10)$ are concyclic.

Three points that do not lie on a single straight line always determine exactly one circle passing through all three; a fourth point is said to be concyclic with the first three if it also lies on that same circle. The standard method is to let the circle be the general equation $x^2+y^2+2gx+2fy+c=0$ (three unknowns $g,f,c$) and substitute the three known points, giving three linear equations that can be solved simultaneously for $g,f,c$ — after which the circle's equation, centre, and radius are all known. An alternative equivalent method sets the unknown centre as $C(h,k)$ and uses the fact that the distances from $C$ to all three points must be equal (since all three are radii of the same circle), giving a pair of equations $CP=CQ$ and $CQ=CR$ that can be solve …