Mathematics - Circle Question with Solution | TestHub

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, $(x-0{)}^2+(y-2{)}^2+\lambda x=0\dots$ (i) Also, it passes through $(-1,0)$. So, $1+4-\lambda =0\Rightarrow \lambda =5$ Eq. (i) becomes, $$\begin{array}{rl}& x^2+y^2-4y+4+5x=0\\ \Rightarrow x^2+y^2+5x-4y+4& =0\end{array}$$ For $x$-intercept, put $y=0$, $$\begin{array}{rlrl}& x^2+5x+4& =0& \\ \Rightarrow & & (x+1)(x+4)& =0\\ \therefore & & x& =-1,-4\end{array}$$