Question

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

Answer 1

The required circle by S + $\lambda$P = 0 is
$x^2+y^2-2x+1+\lambda (x+2y-1)=0$
or $x^2+y^2-x(2-\lambda )+2\lambda y+(1-\lambda )=0$
centre $(-g,-f)is(\frac{2-\lambda }{2},-\lambda )$
$r=\sqrt{g^2+f^2-c}=\sqrt{\frac{(2-\lambda {)}^2}{4}+{\lambda }^2-(1-\lambda )}=\frac{1}{2}\sqrt{5{\lambda }^2}=\frac{\sqrt{5}}{2}|\lambda |$
Since the circle touches the line 2x - y + 3 = 0 therefore perpendicular from centre is equal to
radius $\left|\frac{2.\left(\right(2-\lambda \left)/2\right)-(-\lambda )+3}{\sqrt{5}}\right|=\frac{\left|\lambda \right|}{2}\sqrt{5}\therefore \lambda =\pm 2$
Putting the values of $\lambda$ in (i) the required circles are
$x^2+y^2+4y-1=0\Rightarrow x^2+y^2-4x-4y+3=0$