Suppose there exist a point $(x_{0}, y_{0})$ on the circle $(x-r-1)^2+y^2=r^2$ satisfying $4x_{0}-y_{0}^2\leq0$. Then minimum value of $|r|$ is

Question

Suppose there exist a point $(x_{0}, y_{0})$ on the circle $(x-r-1)^2+y^2=r^2$ satisfying $4x_{0}-y_{0}^2\leq0$. Then minimum value of $|r|$ is

My Approach:

Put given point $(x_{0},y_{0})$ in the given equation of circle $(x-r-1)^2+y^2=r^2\implies$

$(x_{0}-r-1)^2+y_{0}^2=r^2\implies y_{0}^2=r^2-(x_{0}-r-1)^2$

Replaced value of $y_{0}^2$ obtained from above step in $4x_{0}-y_{0}^2\leq0$.

Then I obtained $x_{0}^2-2(r-1)x_{0}+(2r+1)\leq 0$

Now if I do discriminant $D=b^2-4ac\geq0$ I obtain result But why $D\geq0$ is must in this question?

Answer 1

For the simple reason that a discriminant which is negative gives complex non-real values of $x_0$.

When referencing the discriminant you are, more broadly, using an aspect of the quadratic formula, which has a $\sqrt D$ term; if $D < 0$, then you get complex numbers as solutions.

While within a very restrictive framework with very narrow constraints on $x_0,y_0$ you might be able to work with a few complex numbers, broadly this question seems most clearly concerned with $x,y,x_0,y_0,r$ which live in the real numbers.