Are they tangent to each other? How can I show that these two equations are tangent to each other?
$$x-2y=10$$
$$x^2+y^2=20$$
When I sketched and plotted these equations it does show that they are tangent to each other, but I can't show it algebraically.
 A: Hint
Let's find the ordinate of the intersection(s)
$$20=(2y+10)^2+y^2\iff5y^2+40y+80=0$$
$\implies5(y+4)^2=0$
Alternatively
Use the fact:
The perpendicular distance to a tangent from the center $=$ radius
A: The equations only have one solution, $(2,-4)$, so the curves have one intersection point, so they must be tangent (because one is a circle and one is a line).
To find this solution, we can substitute.
$$x = 2y + 10$$
$$(2y+10)^2 + y^2 = 20$$
$$(4y^2 + 40y + 100 + y^2 = 20)$$
$$5y^2 + 40y + 80 = 0$$
Solving this quadratic gives $y=-4$. Substitute back into the other equation to obtain $x=2$.
A: Hints: the line meets the circle at the point $(2,-4)$ as seen by solving the two equations together. The slope of the line is $-\frac 1 2$. To find the slope of the tangent line to the circle at $(2,-4)$ you have to find the derivative using the equation $y =-\sqrt {2-x^{2}}$.  (minus sign because $y<0$ near the point $(2,-4)$). If you carry out this differentiation yo will see that the slope is again $-\frac 1 2$. 
A: HINT
The first depicts a line in $\mathbb{R^2}$, the second a circle.
If you solve the system of the two equations you will be able to deduce-based on the presense and/or number of solutions-which of the three holds:
$i)$ No common points.
$ii)$ A sinlge common point (which means the line is tangent to the circle).
$iii)$ Two common points (the line crosses the circle).
A: The second equation is that of a circle centered at the origin. A line is tangent to a circle iff the distance of the line from the circle’s center is equal to the circle’s radius. Using a standard formula, this distance is $${10\over\sqrt{1^2+2^2}} = 2\sqrt 5.$$ The circle’s radius is $\sqrt{20}=2\sqrt5$, so the line is indeed tangent to the circle.
