The circle passing through the point (-1,0) and touching the Y...

Equation of circle passing through a point $\left(x_1, y_1\right)$ and touching the straight line $L$, is given by $\left(x-x_1\right)^2+$\left(y-y_1\right)^2=\lambda$L=0$\therefore$Equation of circle passing through$(0,2)$and touching$x=0$. Now,$(x-0)^2+(y-2)^2+\lambda x=0 \ldots$(i) Also, it passes through$(-1,0)$. So,$1+4-\lambda=0 \Rightarrow \lambda=5$Eq. (i) becomes,$$\begin{aligned}& x²+y²-4 y+4+5 x=0 \\ \Rightarrow \quad x²+y²+5 x-4 y+4 & =0 \end{aligned}$$For x$-intercept, put $y=0$,$$\begin{array}{rlrl}& x²+5 x+4 & =0 \\ \Rightarrow & & (x+1)(x+4) & =0 \\ \therefore & & x & =-1,-4 \end{array}$$