If a circle and parabola touch each other and also have common root then what's the relationship between their coefficient? Actually this question is from physics (projectile motion) but i believe its related more to maths here equation of parabola
$$y=ax-5x^2-5(ax)^2$$
And that of circle
$$x^2+y^2=(a/5(1+a^2))^2$$ where $a=\tan \theta$ link to original question
https://physics.stackexchange.com/q/562407
Now they touch each other also they share a common root now we have to find the values of a
Now this seems nonsense but if we add slider on graph to a then they seem to meet the required condition at $theta$ approximately equal to $73^\circ$ also no boundattion for initial velocity so ignore the $40 \space m/s$ velocity in graph we can take it $1\ m/s$ also
$\theta$" />
But how do we prove that when $\theta=(\approx)73^\circ$ the circle and parabola meet the required condition of touching each other and having a common root as shown in graph please help me
 A: Let $r = \dfrac{a}{5(a^2+1)}$ be the radius of the circle (the way you wrote it is a bit ambiguous, but this is what you need for the parabola and circle to intersect at $x=r, y=0$).
The resultant of $x^2 + y^2 - r^2$ and $y - (a x - 5 (1+a^2) x^2)$ with respect to $y$ is
$${\frac { \left(  \left( 5\,{a}^{2}+5 \right) x-a \right) 
 \left(  \left( 125\,{a}^{6}+375\,{a}^{4}+375\,{a}^{2}+125 \right) {x}
^{3}+ \left( -25\,{a}^{5}-50\,{a}^{3}-25\,a \right) {x}^{2}+ \left( 5
\,{a}^{2}+5 \right) x+a \right) }{ 25 \left( {a}^{2}+1 \right) ^{2}}}
$$
This must be $0$ where the two curves intersect.  The first factor in the numerator is $0$ at $x = r$, so the other factor in the numerator gives the other intersection.  Now for the curves to be tangent at that point, we want this to have discriminant $0$ with respect to $x$.  That discriminant turns out to be
$$ (62500 (a^4-11 a^2-1)) (a^2+1)^6$$
so the value of $a$ must be the (positive real) root of $a^4 - 11 a^2 - 1$, which is
$$ \sqrt{(11 + 5 \sqrt{5})/2} \approx 3.330190676 $$
This is $\arctan(\theta)$ where $\theta \approx 1.279079822$ radians or $73.28587546$ degrees.
