Equation of circle with A(1, 1), B(2, 2) as ends of diameter is
(x – 1) (x – 2) + (y – 1) (y – 2) = 0 and equation to line AB is x – y = 0.
Hence any circle through A and B is
(x – 1) (x – 2) + (y – 1) (y – 2) + λ(x – y) = 0 ………. (1)
If this C lies in 1st quadrant, it will not cut x or y axis in real point
∴
x2 – 3x + 2 + 2 + λx = 0 and 2 +
y2 – 3y + 2 – λy = 0 will not have real roots on solving we get –1
≤ λ ≤ 1 radius of (1) is
√ will be greatest when λ = ± 1.
Its centre is (2, 1) or (1, 2)