The circle passing through (1,2) and touching the axis of x at (3,0) also passes through the point

Let the centre of the circle be (a;b). Radius will be equal to b, as x-axis is the tangent.
So, the equation of the circle is (x-a)^2 + (y-b)^2 = b^2
As this passes through (3;0), (a-3)^2 + b^2 = b^2 , giving a = 3.
It also passes through (1;2), so, (3-1)^2 + (b-2)^2 = b^2
i.e. 8 - 4b = 0
i.e b = 2
So, the equation of the circle is (x-3)^2 + (y-2)^2 = 4

Shaik Aasif