Question

Consider a family of a circle which are passing through the point (-1,1) and are tangent to x-axis, If (h,k) are the co-ordinate of the centre of the circles, then the set of value of K is given by the interval.

Answer 1

Circle touching $x$-axis and having $(h,k)$ as centre-

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

Given that the circle is passing through $(-1,1)$-

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

$(h^2+1+2h)+(k^2+1-2k)=k^2$

$h^2+2h-2k+2=0$

$D\ge 0$

$\Rightarrow {\left(2\right)}^2-4\times 1\times (2-2k)\ge 0$

$\Rightarrow 8k-4\ge 0$

$\Rightarrow k\ge \frac{1}{2}$

Hence the correct answer is $k\ge \frac{1}{2}$.