A family of circles pass through point (-1,1) and tangent to X axis if (h,k) are coordinates of centre of circles then find range of values of k.

Question

I am unable to approach the question sorry for failing to attempt. Any help appreciated.

Answer 1

Being tangent to the $x$ axis and also passing through $(-1,1)$ requires that the circles be on the positive $y$ axis only. Therefore we know already $0<k$.

But a circle centred on a tangent must have radius zero, so it will never reach as high as $y=1$. In fact we need $0.5\leq k$ as any smaller value forces a radius less than 1 and therefore a circle that cannot pass through the known point while also being tangent to the axis.

For any greater $k$ we could use simple trigonometry and find the required $h$ to make the circle meet both conditions, so there is no upper bound.

Thus altogether, $0.5\leq k<\infty$

Answer 2

$$(x-h)^2+(y-k)^2=r^2$$

Put $y=0\implies x-h=\sqrt{r^2-k^2}$

For tangency, $r^2=k^2$

$$\implies(x-h)^2+(y-k)^2=k^2$$

Now put $(-1,1)$

$$(-1-h)^2+(1-k)^2=k^2\iff2k=2+h^2+2h=1+(h+1)^2\ge1$$

Answer 3

$k>0$ and the smallest vertical diameter is between $ (-1,0),(-1,1)$ at $h=-1$. All other circles tangential to x-axis and above it, having $ y \ge 0$.

The tangent circle radius grows unbounded above x-axis,so,

$$ \frac12\le k < \infty. $$