elder warren
Yo - I'm trying to find m so that the line y=mx-13 is tangent to (x-8)^2+(y+2)^2=16
Here is my idea of how I'd solve it:
I substitute mx-13 into the circle equation as y
I simplify, make into quadratic
I plug it into the quadratic formula
I make delta zero so I only have one solution for m
I find m
Here is what I have so far (I assume I messed up somewhere though don't know where):
y=mx-13
(x-8)^2+(y+2)^2=16
(x-8)^2+(mx-13+2)^2=16
(x-8)^2+(mx-11)^2=16
x^2-16x+64+m^2x^2-22mx+121=16
x^2-16x+169+m^2x^2-22mx=0
a: 1+m^2
b: -16-22m
c: 169
(-16-22m)^2-4+4m^2169=0
484m^2-32m+256-4+4m^2169=0
1160m^2-32m+252=0
(no real roots?)
I hope this makes sense, thanks :)
strong mica