Clever coordinate geometry

Label points $O(0,0), P(m,n), Q(m,0), R(5,0).$ Then since $PR$ is tangent to the given radius $2$ circle, we have that $\Delta OPR$ and $\Delta OQP$ are similar right triangles. By virtue of this similarity we have that

$\dfrac{OQ}{OP} = \dfrac{OP}{OR} \Longrightarrow \dfrac{m}{2} = \dfrac{2}{5} \Longrightarrow \boxed{m = \dfrac{4}{5}}.$

To solve the problem you nave to find the angular coefficient of the line.

$y=ax+b$

Using the point (5,0), we know that 0=5a+b, so b=-5a. Now we have $y = a(x-5)$

Now using in the circumference equation we have:

$x^2 + y^2=4 \longrightarrow x^2 + [a(x-5)]^2=4$

$(a^2 +1)x^2 -(10a^2)x +25a^2 -4=0$

Because the line is tangent to circumference, the equation must have just one solution and the discriminant of the quadratic must be 0.

$100a^4 -4(a^2 +1)(25a^2-4)=0 \longrightarrow 100a^4 -100a^4 +16a^2 -100a^2 +16=0 \longrightarrow a^2=4/21$

Using that

$m= -\frac{b}{2a} = -\frac{-10a^2}{2(a^2+1)} = \frac{10(\frac{4}{21})}{2(\frac{4}{21} +1)} = \boxed{\frac{4}{5}}$

T=0 Therefore xm+yn=4 since it passes through (5,0) m= 0.8