Let there be two curves $S:(x-\sqrt{7})^2+y^2=R^2$ and $S':\dfrac{x^2}{16}+\dfrac{y^2}{9}=1$, then what is the value of $R$ such that $S$ and $S'$ touch at exactly one point?
substituting $y^2$ from $S$ into $S'$ we get $$\dfrac{x^2}{16}+\dfrac{R^2-(x-\sqrt{7})^2}{9}=1$$
$$\frac{1}{144}( 7 x^2 - 32 \sqrt{7} x +256-16R^2)=0$$
Now since the curves should only touch at exactly one point, the discriminant of this quadratic must be $0$,
$$\Delta=\left(32\sqrt{7}\right)^{2}-4\cdot7\cdot\left(256-16R^{2}\right)=0\implies R=0$$ but this is wrong as the given answer is $R=4\pm \sqrt{7}$ which is true
So why is discriminant equals zero not giving the right answer here?
The circle is centered at a point (focus of the ellipse) lying on the major axis of the ellipse, which is horizontal. The unique common point of the two curves lies necessarily on the same line. Therefore, $y=0.$
Uploading into $S'$ gives $$x^2=16\quad \text{or}\quad x=\pm 4.$$ Substituting into $S$ we get $$(\pm 4-\sqrt7)^2+0^2=R^2$$ or equivalently $$R=4\pm \sqrt7$$