How to find the equation of the circle which touches $y$ axis at $(0,3)$ and cuts a chord of length $8$ on the $x$ axis?
It should look like this:
Since the circle touches $y$ axis at $(0,3)$, its center has $y$-coordinate $3$. So the equation of the circle is of the form $(x-r)^2+(y-3)^2=r^2$.
How can I proceed further by using the fact that the circle passes through $(a,0)$ and $(b,0)$ with $b-a=8$? Anyway this would be very long. Is there some alternative?