Equation of the line through the point $(\frac{1}{2},2)$ and tangent to the parabola $y=\frac{-x^2}{2}+2$ and secant to the curve $y=\sqrt{4-x^2}$ Find the equation of the line through the point $(\frac{1}{2},2)$ and tangent to the parabola $y=\frac{-x^2}{2}+2$ and secant to the curve $y=\sqrt{4-x^2}$

Let the required line is tangent to the parabola at the point $(x_1,y_1)$.It passes through $(\frac{1}{2},2)$.Its equation is $y-2=-x_1(x-\frac{1}{2})$.
This line is also secant to the the curve $y=\sqrt{4-x^2}$.
I solved $y=\sqrt{4-x^2}$ and the line $y-2=-x_1(x-\frac{1}{2})$.
I am stuck here.

HINT:
The equation of any straight line passing through $(1/2,2)$ is $$\dfrac{y-2}{x-1/2}=m\iff y=mx+2-m/2$$ where $2m$ is the gradient.
Let us find the intersection of this line with the given parabola.
$$mx+m-\dfrac12=2-\dfrac{x^2}2\iff x^2+2mx+2m-5=0$$ which is a quadratic equation  in $x$  whose each of the two roots represents the abscissa of  intersection. For tangency, both roots must be same.
