The following is a problem from the chapter "Circles":
Consider a family of circles which are passing through the point $(−1, 1)$ and are tangent to $x$-axis. If $(h, k)$ are the co-ordinates of the centre of the circles, then the set of values of $k$ is given by the interval:
(A) $0 < k < 1/2$
(B) $k\ge1/2 $
(C) $− 1/2 \le k \le 1/2 $
(D) $k \le1/2 $
My Approach:
It is given that the family of circles are passing through a fixed point $(-1,1)$ which is above the $x$-axis. It is also given that the $x$-axis is the tangent for all the circles, or in other words all circles touch the $x$-axis at one point. We are asked to find the set of values taken by $k$ which is the $y$ coordinate of the centre of the family of circles. Since $x$-axis is tangent to all the circles, the value of $k$ is same as the radius of the circle it represents. Now the question condenses to finding the range of radii of the circles under the given condition.
There is no upper limit on the value of the radius but there is a lower limit because, there must always be contact with the point $(-1,1)$. Thus from this fact and thinking graphically, we can say that the least possible value of radii is $1/2$. It can't be lesser than this since if it does so, the contact with the fixed point in the question will be lost.
Thus, I concluded that the set of values taken by $k$ is $k\geq1/2$ (option (B)) and the answer obtained in correct with respect to my book!
But, I think my way of solving involves some kind of graphical reasoning (visualisation) which may be prone to errors in some cases. Even though I solved this problem correctly and quickly, I wish to know other formal solutions for this, which might be useful in other circumstances.
Thank you in advance.
As you noticed, the radius is $k$, so the equation of the circle is given by $$(x-h)^2+(y-k)^2=k^2$$ Since this circle passes through $(-1,1)$, we get $$(-1-h)^2+(1-k)^2=k^2,$$ i.e. $$h^2+2h+2-2k=0\tag1$$
Now, considering the discriminant helps.
The real number $h$ satisfying $(1)$ exists if and only if $$2^2-4\times 1\times (2-2k)\ge 0,$$ i.e. $$\color{red}{k\ge \frac 12}$$