Minimum value of $\displaystyle \left(x_{1}-x_{2}\right)^2+\left(12-\sqrt{1-x^2_{1}}-\sqrt{4x_{2}}\right)^2.$ The Minimum value of $\displaystyle \left(x_{1}-x_{2}\right)^2+\left(12-\sqrt{1-x^2_{1}}-\sqrt{4x_{2}}\right)^2\;,$ Where $x_{1}\;,x_{2}\in \mathbb{R}$.
$\bf{My\; Trial}::$ Here we have to find Distance between the point $\displaystyle A\left(x_{1}\;,12-\sqrt{1-x^2_{1}}\right)$ and 
point $\displaystyle B\left(x_{2}\;,2\sqrt{x_{2}}\right).$
But i did not understand in which the curve these $2$ points lies.
Help me Thanks

One point is in the circle with radius 1 and with center at (0,12) and the other point is in a parabola $x= y^2/4$.  I used EXCEL and found that the shortest distance and the minimum  value is 7.944277 and the points are (0.45,11.06971) on the circle and (4.0,4.0). As the other responder said, you could use calculus to find the minimum.
The shortest distance for any point outside of a circle with coordinates (x,y) is the normal line to the center of the circle.
Thus $D = \sqrt{x^2+(y-12)^2}$ where $y = 2\sqrt{x}$
$\frac{dD}{dx} = \frac{2x + 2(2\sqrt{x} - 12).\frac{1}{\sqrt{x}}}{2(D)} = 0$
$x + 2\sqrt{x} -12 = 0$
$(\sqrt{x}(x+2))^2 = 144$
$x^3+4x^2+4x-144 = 0$
A casual observation will let you kmow x=4 is a root for this cubic function.
The other roots are non-real.
Thus the point is (4,4) in the parabola.
Now the normal from that point will have a slope $=>(m_1m_2) = -1 => \frac{1}{2}m_2 = -1=> m_2 = -2$ where $m_1 => 2y(\frac{dy}{dx}) = 4=> \frac{dy}{dx} =  m_1 =\frac{1}{2}$
The normal equation =>$ y-4 = -2(x-4)$
Now substitite the y would get from the circle where the normal would intersect.
$12-\sqrt(1-x^2) -4 = -2x+8 => 8-\sqrt{(1-x^2)} = -2x+8 => 1-x^2 =4x^2=>5x^2 = 1=> x^2 =\frac{1}{5} => x = \sqrt{\frac{1}{5}}$
Thus the other point is $(\sqrt{\frac{1}{5}},11.105572)$
Minimum$ D = \sqrt{(\sqrt{\frac{1}{5}}-4)^2+(11.105572-4)^2}$
$ = \sqrt{(12.622289+50.489153} = 7.944271$ precisely.
