Finding tangent of a circle given equation of circle and tangent with unknown gradient Given the equation of the circle $(x-4)^2 +(y+1)^2 = 4$ and equation of a tangent to the circle $y = mx$, I am supposed to find the values of $m$. I've tried various methods, such as equating the two equations and finding the roots, but I cannot $m$ in isolation. Can anyone give a little intuition or guidance? Thanks!
 A: As its a tangent , distance from centre of the circle=radius of the circle. Thus we have $$|\frac {-1-4m}{\sqrt {1+m^2}}|=2$$ ie $(1+4m)^2=4+4m^2$. Solving this quadratic we have the answer.
A: $y = mx$ is true for any point on the line (that is, after all, what defines the connection between the equation for a line and the line itself). Specifically, it is true for whatever points the line and the circle have in common. The other equation is true for any point on the circle, and specifically, it is true for any point of the intersection between line and circle. So on the intersection the two equations are both true (and where the equations are both true is on the intersection), which means you have a set of two equations describing exactly the $x$ and $y$ coordinates of the intersection.
Start solving those equations, and keep in mind that a tangent of a circle intersects the circle in exactly one point, not two or none, which means that the equations must have exactly one solution, not two or none. There are only a limited number of values of $m$ that makes this happen.
A: Just solve $$\frac{|4m+1|}{\sqrt{m^2+1}}=2.$$
