The circle passing through the point $(-1,0)$ and touching the $Y$-axis at $(0,2)$, also passes through the…
The circle passing through the point $(-1,0)$ and touching the $Y$-axis at $(0,2)$, also passes through the point
$\left(-\frac{3}{2}, 0\right)$
$\left(-\frac{5}{2}, 2\right)$
$\left(-\frac{3}{2}, \frac{5}{2}\right)$
$(-4,0)$
Solution
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}
& x^2+y^2-4 y+4+5 x=0 \\
\Rightarrow \quad x^2+y^2+5 x-4 y+4 & =0
\end{aligned}
$
For $x$-intercept, put $y=0$,
$
\begin{array}{rlrl}
& x^2+5 x+4 & =0 \\
\Rightarrow & & (x+1)(x+4) & =0 \\
\therefore & & x & =-1,-4
\end{array}
$