I have the following point $A(-1;1)$ and the line $s:x-2y+3=0$. Consider the general equation of the circumference:
$$x^2+y^2+ax+by+c=0.$$
I applied the following transformation rule:
$$ x^2 \to x \cdot x_0, \quad y^2 \to y \cdot y_0, \quad x \to \frac{x+x_0}{2}, \quad y \to \frac{y+y_0}{2} $$
and using for $x_0=-1$ and $y_0=1$ I obtain:
$$ x\left(\frac{a}{2}-1\right)+y\left(\frac{b}{2}+1\right)-\frac{a}{2}+\frac{b}{2}+c=0 $$ Now if impose that the coefficients of this generic line must be equal to the coeffcients to my line $s$, I obtain the following system:
\begin{cases} \frac{a}{2}-1=1,\\ \frac{b}{2}+1=-2,\\ -\frac{a}{2}+\frac{b}{2}+c=3 \end{cases} with just ONE solution for $a=4,b=-6,c=8$. Why I have this result? I think that I shoud have a bundle of circumferences that are tangent in that point with the line $s$. Why I obtain this result withi this method? Thanks to all.
The coefficients of the two lines mustn’t be equal, only proportional for them to be identical, i.e. $$\frac a2-1 = t \\ \frac b2 +1 =-2t \\ -\frac a2 +\frac b2 +c =3t $$ for some $t \ne 0$, giving you an infinite number of such circles.