Hello There,

The general equation of a circle is given as :

x^2 + y^2 +2gx + 2fy + c = 0

Now if the circle passes through these three points then obviously these three points satisfy the equation of circle.  So

For point P , we have

1+1+2g+2f+c=0       (eqn.1)

for point Q, we have

4+1+4g-2f+c=0        (eqn 2)

for point R, we have

9+4+6g+4f+c =0      (eqn 3)

Solving these three equation we get that ;

g = -5/2    f = -1/2    and  c = 4

So the equation of circle is given by, putting the values of f, g and c in the general eqn of circle.

eqn = x^2 + y^2 - 5x - y + 4=0

Hope this helps.