<s>[INST] Distance between points - equation of a line I have worked on this particular example: 
The distance between the point $M_1(3,2)$ and $A$ is $2\sqrt5$ and the distance between the point $M_2(-2, 2)$ and $B$ is $\sqrt5$. Come up with a equation for the line going through $A$ and $B$. 
I have tried playing around with the equations for the circles with centers in $M_1$ and $M_2$ respectively (the radii being the distance between $M_1$ and $A$, that is the distance between $M_2$ and $B$ for the other circle), but I am stuck. 
Would appreciate any help.
Thanks in advance. 
P.S. I am probably wrong but, isn't there infinitely many solutions since with the data I'm given I am basically being asked to come up with a particular equation for a line connecting any two points on those circles ? 
 [/INST] Point on the circle with radius $2\sqrt{5}$ and center $(3,2)$ is at:
$$ A = (2\sqrt{5} \cos(a) + 3, 2\sqrt{5} \sin(a) + 2)$$
The other circle's point is at:
$$ B = (\sqrt{5} \cos(b) - 2, \sqrt{5} \sin(b) + 2) $$
Then slope of the line is:
$ m = \frac{B_y - A_y}{B_x-A_x} = \frac { \sqrt{5} \sin(b) + 2 - (2\sqrt{5} \sin(a) + 2) }{ \sqrt{5} \cos(b) - 2 - (2\sqrt{5} \cos(a) + 3)} = \frac{\sqrt{5}(\sin(b)-2\sin(a))}{\sqrt{5}(\cos(b)-2\cos(a))-5}$ 
Then the equation of the line is:
$$ y-A_y = m(x-A_x)$$
$$ y-(2\sqrt{5} \sin(a) + 2) = m(x-(2\sqrt{5} \cos(a) + 3)) $$
$$ y-(2\sqrt{5} \sin(a) + 2) = \frac{\sqrt{5}(\sin(b)-2\sin(a))}{\sqrt{5}(\cos(b)-2\cos(a))-5} (x-(2\sqrt{5} \cos(a)+3)) $$
</s>