Question

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

Answer 1

Let the centre of the circle be $(h,2)$ then radius$=\left|h\right|$

Equation of the circle becomes ${(x-h)}^2+{(y-2)}^2=h^2$

As it passes through $(-1,0)$

$\Rightarrow {(-1-h)}^2+{(0-2)}^2=h^2$

$\Rightarrow 1+h^2+2h+4=h^2$

$\Rightarrow 2h+5=0$ or $h=\frac{-5}{2}$

$\therefore$Distance of the centre from $(x,y)$ is

${(x+\frac{5}{2})}^2+{(y-2)}^2=\frac{25}{4}$

Let us check with option$\left(a\right)$

Distance of the centre from $(\frac{-3}{2},0)$ is

${(\frac{-3}{2}+\frac{5}{2})}^2+{(0-2)}^2=1+4\ne \frac{25}{4}$

Let us check with option$\left(b\right)$

Distance of the centre from $(\frac{-5}{2},2)$ is

${(\frac{-5}{2}+\frac{5}{2})}^2+{(2-2)}^2=0\ne \frac{25}{4}$

Let us check with option$\left(c\right)$

Distance of the centre from $(\frac{-3}{2},\frac{-5}{2})$ is

${(\frac{-3}{2}+\frac{5}{2})}^2+{(\frac{-5}{2}-2)}^2=1+\frac{81}{4}=\frac{85}{5}\ne \frac{25}{4}$

Let us check with option$\left(d\right)$

Distance of the centre from $(-4,0)$ is

${(-4+\frac{5}{2})}^2+{(0-2)}^2=1+\frac{81}{4}=\frac{9}{4}+4=\frac{9+16}{4}=\frac{25}{4}$

$\therefore$ it passes through $(-4,0)$