1. Centre = mid-point of P and Q, say M = (h, k) = ( (1-a)/2, (5+a)/2 ). Radius = r = MP Equation of circle is (x-h)^2 + (y-k)^2 = r^2 2. Method 1: Since (-2, 1) lies on the circle, it is in quadrant 2. Hence, its equation is (x+r)^2 + (y-r)^2 = r^2, i.e. x^2+y^2+2rx-2ry+r^2=0. Putting (-2, 1) into it gives r=1, 5. Method 2: Let its equation be C: x^2 + y^2 + Dx + Ey + F = 0 Since it touches coordinate axes, C has unique solution at x=0 and y=0, i.e. x^2+Dx+F=0 and y^2+Ey+F=0 have unique solution, i.e. D^2-4F=E^2-4F=0, i.e. E = ±D, F=D^2/4. Together with (-2,1) lies on C, E=D is rejected, and this gives the same answer. 3. Let the mid-points of A and B, B and C, and A and C be D, E and F respectively; G(h,k) be the centre. Since the line from centre perpendicular to chord bisects chord, we have DG is perpendicular to AB, EG is perpendicular to BC and FG is perpendicular to AC. Since AB is a vertical line, DG is horizontal, i.e. they have same y-coordinate, i.e. k=(3+5)/2. Also, we have slope of EG times slope of BC = -1, etc. and we can then solve for h which is h=-1. The radius = r = AG. Finally, the equation is (x-h)^2 + (y-k)^2 = r^2. 4. Let G(h, k) be the centre and h=±2. Since G lies on x+2y-4=0, k can be found.

Its equation is (x-h)^2 + (y-k)^2 = r^2.