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# The circle passing through the point $(-1,0)$ and touching the y-axis at $(0,2)$ also passes through the point .

$\begin{array}{1 1}(A)\;(\frac{-3}{2},0) \\(B)\;(\frac{-5}{2},2) \\(C)\;(\frac{-3}{2},\frac{5}{2}) \\(D)\; (-4,0) \end{array}$

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A)
Let the centre of the circle be $(h,2)$ then radius $=|h|$
equation of circle becomes $(x-h)^2+(y-2)^2=h^2$
As it passes through $(-1,0)$
=> $(-1-h)^2+4=h^2$
=> $h= \large\frac{-5}{2}$
Centre $\bigg( \large\frac{-5}{2},$$2\bigg)$
and $r= \large\frac{5}{2}$
Distance of centre from $(-4,0)$ is $\large\frac{5}{2}$
$\therefore$ It lies on the circle.