Let x=0, y 2-2by+b 2+a 2-r 2 = 0.

| y 1-y2 | =√[(y 1+y2)2-4y 1 y2]= 2 √( R2-a2)，r 2 = a 2+ 1 ①。

Let y=0, x 2-2ax+a 2+b 2-R2 = 0.

| x 1-x2 | =√[(x 1+x2)2-4x 1x 2]= 2 √( R2-B2)，r 2 = 2b 2 ②。

2b 2-a2 =1is obtained from ① and ②.

And because the distance from P(a, b) to the straight line x-2y = 0 is √5/5.

D=|a-2b|/√5=√5/5, that is, A-2b = 1.

To sum up, we can get a= 1, b= 1 or a =- 1, b =- 1.

So r 2 = 2b 2 = 2.

The equation of a circle is (x+ 1) 2+(y+ 1) 2 = 2 or (x- 1) 2+(y- 1) 2 = 2.