Verifying that the equation of a line is a tangent Consider the circle: $x^2 + y^2 = 2$ The question says to verify if the equation $y = 2x + 1$ is a tangent of the circle or not.
I was told to solve the equation but I still don't understand how that'll help in the verification. 
 A: HINT
We need to check that the system


*

*$x^2+y^2=2$

*$y=2x+1$


has exactly one solution.
Then plug in $y$ from the second in the first equation to obtain a quadratic equation in $x$.
A: A line may pass through a circle, touch it, or miss it altogether. In each case, how many points are there that are on both the circle and the line? Which case is the tangent? Any such point satisfies both of the equations. How many solutions do you need? How many solutions are there? Solve the eqations simultaneously and find out.
A: Distance of line $Ax+By +C=0$ to the origin is given by:
$d:= |\dfrac{A0+B0+C}{\sqrt{A^2+B^2}}|$.
With $2x-y+1=0$ we get:
$d=\dfrac{1}{√5}$. 
Radius of the circle is $√2$.
Is the line a tangent to the circle ?
A: A slight variation on Peter's.
On the line y=2x+1, how far is the point of closest approach to the center of the circle, i.e. the origin?
The slope of a line from the center to point of closest approach is -1/m where m is the slope of the given line. Here m=2, so -1/m= -1/2.
Find where (-1/2)*x =y intersects y=2x+1. Given that x and y, find the distance to the origin. If the point of closest approach is within the circle or outside the circle, it's not a tangent. It's only a tangent if the distance of the points of closest approach is equal to the radius. 
