The Circle $$x^2+y^2-4x=0$$ is cut by a line $AB$ at two points. If $A$,$B$ and two other points $C(1,0)$ and $D(0,1)$ are Concyclic, Then which of the Following points are not contained by the line.
$1.(1,1)$
$2.(-2,2)$
$3.(\frac{7}{4},\frac{3}{4})$
My Try: Equation of any Circle Passing through Circle and Line Intersections is
$$S+kL=0$$ i.e., $$x^2+y^2-4x+k(ax+by-1)=0$$ which passes through $C(1,0)$ and $D(0,1)$. we get two equations
$$a=1+\frac{3}{k}$$ and $$b=1-\frac{1}{k}$$
But how can we find $k$..Help needed
Circle($C_1$) is: $$(x-2)^2+y^2=4$$ Line($L_1$) could be taken as: $$y=mx+c$$ For the circle passing through intersection of $C_1$ and $L_1$ could be taken as: $$C_1+\lambda L_1=0\implies (x-2)^2+y^2-4+\lambda(y-mx-c)=0$$ This passes through $(1,0)$ and $(0,1)$ both: $$1+\lambda(1-c)=0\text{ and }-3+\lambda(-m-c)=0$$ So: $$\require{cancel}\frac{\cancel{\lambda}(1-c)}{\cancel{\lambda}(-m-c)}=\frac{-1}{3}\implies 3-3c=m+c\implies c=\frac14(3-m)$$ So: $$y=mx+\frac14(3-m)\implies 4y=4mx+3-m\implies \left(y-\frac34\right)=m\left(x-\frac14\right)$$ Clearly the line passes through $(1/4,3/4)$ and I have tested it on an app and all points seem to be lying on this line.
Graphs:
Legend:
$$\color{red}{(x-2)^2+y^2=4}\quad\color{blue}{y=mx+c}\quad\color{black}{(x-2)^2+y^2-4+k(y-mx-c)=0} \\\color{red}{(0,1)}\quad\color{green}{(1/4,3/4)}\quad\color{purple}{(1,1)}\quad\color{orange}{(-2,2)}\quad\color{black}{(7/4,3/4)}$$
$$\large m\approx0.33\text{ [actually it should be exactly }1/3]$$
$$\large m\approx0\text{ [actually it should be exactly }0]$$
$$\large m\approx-0.56\text{ [actually it should be exactly }-5/9]$$