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Theory [Equation of a circle passing through a point and a tangent not on that point]

Consider that the equation of the circle is \(g(x,y)\) and \(f(x,y)\) as the equation of a straight line. We consider \(g(x,y)\) as a point circle that intersects \(f(x,y)\) at some point say (\(m,n)\). A point circle is basically a point, so it must intersect the line at just a single point. Therefore, since the circle formed from the intersection of a circle and a straight line is given by the A. R. Khalifa rule: $$g(x,y)+k. f(x,y)=0$$ where \(k\) is a constant, we consider that this must be the equation of the circle that we need to find out. Now, there are few things to notice. First, the primary circle we considered i. e. \(g(x,y)\) that it is a point circle, which practically is not. Second, how do we find \(k\)?
Now, the solution is, both the tangent and the circle goes through the point (\(m,n)\), which leads to \(g(m,n)=0\) and \(f(m,n)=0\), that does not help finding \(k\). However, we have another point (\(p,q)\) through which none of the circle \(g(x,y)\) or the straight line \(f(x,y)\) passes through, but the circle \(g(x,y)+k.f(x,y)\) does. Therefore, we can find \(k\) from this equation, which is: $$k=-\frac{g(p,q)}{f(p,q)}$$ Now, it is time to consider \(g(x,y)\). Since, it has a radius of \(0\) and it is centered on \((m,n)\), we have $$g(x,y)=x^2+y^2+2mx+2ny+m^2+n^2=0$$ The circle must be centered on \((m,n)\), since it touches the straight line \(f(x,y)\) on this point. If it is centered anywhere else, it gains a non-zero radius, that is contradictory to the condition that we set up at the beginning. Just as the cases before, the straight line has the form \(f(x,y)=ax+by+c=0\). Therefore, the value of \(k\) finally leads to: $$k=-\frac{p^2+q^2+2mp+2nq+m^2+n^2}{ap+bq+c}$$ Finally the equation of the required circle \(z(x,y)=g(x,y)+k. f(x,y)=0\) is:

$$x^2+y^2+2mx+2ny+m^2+n^2-\frac{p^2+q^2+2mp+2nq+m^2+n^2}{ap+bq+c}(ax+by+c)=0$$