If (a,b) be the centre of the circle and radius=r and clearly the circle passes through (1,0)
then $r^2=(1-a)^2+b^2$
The equation of the circle $(x-a)^2+(y-b)^2=(1-a)^2+b^2$
The gradient of the circle at (x,y) = $\frac{dy}{dx} = \frac{(a-x)}{(y-b)}$
The gradient of the circle at (0,3) =$ \frac{(a-0)}{(3-b)}$
As y-axis is tangent to the circle at (0,3) and its gradient is ∞, so b=3.
(i)The equation of the circle becomes $(x-a)^2+(y-3)^2=(1-a)^2+3^2$. As the circle passes through (0,3), $(0-a)^2+(3-3)^2=(1-a)^2+3^2$ =>a=5.
Or (ii) As the circle passes through (0,3), $r^2=(0-a)^2+(3-b)^2$, but $r^2=(1-a)^2+b^2$ =>a=3b-4 =>a=5
As any intersection of x-axis & the circle, (x,0)
=>$(x-a)^2+(0-b)^2=(1-a)^2+b^2$
=>x=1,2a-1.
So, the other x-intercept is 2(5)-1=9
Alternatively, the circle passes through (0,3), (1,0).
Let the equation of the circle : $x^2+y^2+2gx+2fy+c=0$
9+6f+c=0 and 1+2g+c=0.
Let the expected x-intercept be t, so the third point on the circle (t,0). So, $t^2+2gt+c=0$
So, t,1 are the roots of $s^2+2gs+c=0$
=>t+1=-2g and t.1=c
=>2g=-t-1 and c=t
As 9+6f+c=0, 2f=-$\frac{9+c}{3}$
So, the equation of the circle becomes $x^2+y^2-(t+1)x-\frac{9+t}{3}y+t=0$
The gradient of the circle at (x,y) = $\frac{dy}{dx} = \frac{3(2x-t-1)}{(9+t-6y)}$
The gradient of the circle at (0,3) =-$\frac{t+1}{t-9}$
As y-axis is tangent to the circle at (0,3) and its gradient is ∞, so t=9.
So, the other x-intercept is 9