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.
 A: Let $\left(t,-\frac{t^2}{2}+2\right)$ be a tangent point.
Since $\left(-\frac{x^2}{2}+2\right)'=-x$, we get an equation of the tangent line:
$$y+\frac{t^2}{2}-2=-t(x-t).$$
Now, substitute $x=\frac{1}{2}$ and $y=2$, find a values of $t$ (I got $t=0$ or $t=1$) and choose a value, which you need.
A: 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.
A: Problem
$$
 \color{red}{y = 2 - \frac{x^{2}}{2}}
\tag{1}
$$
$$
 \color{blue}{y = \sqrt{4-x^2}}
\tag{2}
$$


Equation of the line
Find the equation of the tangent line 
$$
 y = m x + b
\tag{3}
$$
tangent to $\color{red}{y(x)}$
We are given the a point 
$$
 p = \left( \frac{1}{2}, 2 \right)
$$
Find the slope, $m$, and the intercept, $b$.
Slope
To be tangent to the red curve, the slope of the line must match the slope of red curve. The slope is of the red curve is 
$$
 \color{red}{y'} = -1
$$
Intercept
The intercept is computed from $(3)$ using the point $p$:
$$
 b = y - m x \qquad \Rightarrow \qquad b = 1 - (-1) \frac{1}{2} = \frac{5}{4}
$$
Solution
The equation for the tangent line (the dashed line below) is
$$
\boxed{ y = -x + \frac{5}{4}}
\tag{4}
$$
Tangent point
Where does the dashed line, $(4)$, touch the red curve, $(1)$? Solve
$$
\begin{align}
  y &= \color{red}{y} \\
%
  -x + \frac{5}{4} &= \color{red}{2 - \frac{x^{2}}{2}} \\
%
 x &= 1
\end{align}
$$
Using $(4)$, we have the tangent point is
$$
  q = \left( 1, \frac{3}{2} \right),
$$
where the dashed line touches the red curve.

Secant points
Where are the two points where the dashed line, $(4)$ intercepts the blue curve, $(2)$? Solve
$$
\begin{align}
 y &= \color{blue}{y} \\
 -x + \frac{5}{4} &= \color{blue}{\sqrt{4-x^2}} \\
\end{align}
$$
The solution is $x = \frac{1}{4} \left(5 \pm \sqrt{7} \right)$. Therefore the two points define the secant chord are
$$  
%
  \frac{1}{4} 
  \left( \left(5 - \sqrt{7}\right), 
  \left(5 + \sqrt{7}\right) + 10 \right), \qquad
%
  \frac{1}{4} 
  \left( \left(5 + \sqrt{7}\right),  
  \left(-5 - \sqrt{7}\right) + 10 \right)
%
$$
