Find the equation of circle which touches $2x-y+3=0$ and pass through the points of intersection of the line $x+2y-1=0$ and the circle $x^2+y^2-2x+1=0$

The correct options are
A

B

Equation family of circle is

$S+\lambda L=0$

$x^2+y^2-2x+1+\lambda (x+2y-1)=0$

$x^2+y^2-x(x-\lambda )+2\lambda y+(1-\lambda )=0$ - - - - - -(1)

centre of circle is (-g,-f)

$=(\frac{2-\lambda }{2},-\lambda )$

And radius of the circle is =$\sqrt{g^2+f^2-c}=\sqrt{{\left(\frac{2-\lambda }{2}\right)}^2+{\lambda }^2-(1-\lambda )}$

$=\frac{1}{2}\sqrt{4+{\lambda }^2-4\lambda +4{\lambda }^2-4+4\lambda }$

$=\frac{1}{2}\sqrt{5{\lambda }^2}=\frac{\sqrt{5}}{2}|\lambda |$

Since, circle touches the line $2x-y+3=0$, therefore perpendicular from the centre should be equal to

radius.

Centre $(\frac{2-\lambda }{2},-\lambda )$

Perpendicular distance from a point to a line is

$\left|2.\frac{\left(\frac{(2-\lambda )}{2}\right)-(-\lambda )+3}{\sqrt{4+1}}\right|=\frac{\sqrt{5}}{2}|\lambda |$

$\frac{|2.\lambda +\lambda +3|}{\sqrt{5}}$

$\left|\frac{5}{\sqrt{5}}\right|=\frac{\sqrt{5}}{2}|\lambda |$

$\sqrt{5}=\frac{\sqrt{5}}{2}|\lambda |$

$|\lambda |=2$

$\lambda =\pm 2$

Substituting the value of $\lambda$ in equation (1)

When $\lambda =2$

$x^2+y^2-x(2-2)+2\left(2\right)y+(1-2)=0$

$x^2+y^2+4y-1=0$

When $\lambda =-2$

$x^2+y^2-x(2+2)+2(-2)y+(1+2)=0$

$x^2+y^2-4x-4y+3=0$

so,option A and B are correct.