Question

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

• A. $(-\frac{3}{2},0)$
• B. $(-\frac{5}{2},2)$
• C. $(-\frac{3}{2},\frac{5}{2})$
• D. $(-4,0)$

Answer 1

The correct option is D $(-4,0)$

Let $(h,k)$ be center of circle.

Circle touches the $y$-axis.

$\therefore$ Radius of circle $=h$

Equation of circle-

${(x-h)}^2+{(y-k)}^2=h^2.....\left(1\right)$

Since the circle passes through $(0,2)$

Therefore,

$h^2+{(2-k)}^2=h^2$

$\Rightarrow {(k-2)}^2=0$

$\Rightarrow k=2$

Given that the circle also passes through $(-1,0)$,

Therefore,

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

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

$h=-\frac{5}{2}$

Substituting the value of $h$ and $k$ in ${eq}^n\left(1\right)$, we get

${(x+\frac{5}{2})}^2+{(y-2)}^2={\left(\frac{5}{2}\right)}^2$

Now, we can find the point from which the circle passes.

As the point $(4,0)$ satisfy the equation of circle.

Thus, the circle will also pass through the point $(-4,0)$.

Hence the correct answer is $\left(D\right)(-4,0)$.