A circle is tangent to the $y$-axis at $y=3$ and has one $x$-intercept at $x=1$. Find the other $x$-intercept 
A circle is tangent to the $y$-axis at $y=3$ and has one $x$-intercept at $x=1$. Find the other $x$-intercept 

Like previously mentioned, I'm not all too familiar with circles. So, I plotted the two points and I do not know the next step. I would guess to find the slope which is $-\dfrac{3}{1}$. But that doesn't seem right. I'm just taking a shot in the dark. If someone can tell me what to do next, or at least how to find the center that would be helpful. Please do not give me the answer.

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
