The circle possessing $y$-axis as its tangent at $(0,2)$ and passing through $(-1,0)$, also passes through
The circle possessing $y$-axis as its tangent at $(0,2)$ and passing through $(-1,0)$, also passes through
$\left(\frac{-3}{2}, 0\right)$
$\left(\frac{-5}{2}, 2\right)$
$\left(\frac{-3}{2}, \frac{5}{2}\right)$
$(-4,0)$
Solution
Let $(h, k)$ be centre of circle.
Circle touches the $y$-axis
$\therefore$ Radius of circle $=\mathrm{h}$
Equation of cicle
$
(\mathrm{x}-\mathrm{h})^2+(\mathrm{y}-\mathrm{k})^2=\mathrm{h}^2 .
$
It passes through $(0,2)$
$
\therefore \mathrm{h}^2+(2-\mathrm{k})^2=\mathrm{h}^2 \Rightarrow \mathrm{k}=2
$
Since, it also passes through $(-1,0)$
$
\therefore(-1,-h)^2+2^2=h^2 \Rightarrow \mathrm{h}=-5 / 2
$
$\therefore$ Equation of circle: $\left(x+\frac{5}{2}\right)^2+(y-2)^2=\left(\frac{5}{2}\right)^2$ Only option (d) $(-4,0)$ satisfied it