<s>[INST] Find the equation of a circle which is tangent to $y$-axis at a given point and cuts a chord of given length on the $x$-axis 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:

My approach:
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?
 [/INST] $\begin{vmatrix}
x^2+y^2&x&y&1\\
0^2+3^2&0&3&1\\
1^2+0^2&1&0&1\\
9^2+0^2&9&0&1\\
\end{vmatrix}
=
\begin{vmatrix}
x^2+y^2&x&y&1\\
9&0&3&1\\
1&1&0&1\\
81&9&0&1\\
\end{vmatrix}=  
24(x^2+y^2)-240x-144y+216=  
x^2+y^2-10x-6y+9=0$
</s>