[ASK] Tangent of a Circle (Again)

[Monoxdifly]

If the line x + my = 1 is a tangent of the circle \(\displaystyle x^2+y^2-4x+6y+8=0\), the value of m is ...
A. -2
B. \(\displaystyle \frac{1}{4}\)
C. \(\displaystyle \frac{1}{4}\)
D. 3
E. 4

Looking at the circle's equation, the center is (2, -3) and the radius is \(\displaystyle \sqrt5\). If I know the coordinate where the line meet the circle I think I can solve this by myself, but I don't know how. Should I substitute x = 1 - my to the circle equation, or is there a simpler way?

 

[MarkFL]

...Should I substitute x = 1 - my to the circle equation, or is there a simpler way?

I think that's what I would do...and then equate the discriminant of the resulting quadratic in $y$ to zero. :)

 

[Monoxdifly]

Fine... Let's try:
\(\displaystyle x^2+y^2-4x+6y+8=0\)
\(\displaystyle (1-my)^2+y^2-4(1-my)+6y+8=0\)
\(\displaystyle 1-2my+m^2y^2+y^2-4+4my+6y+8=0\)
\(\displaystyle (m^2+1)y^2+(6+2m)y+5=0\)

D = 0
\(\displaystyle b^2-4ac=0\)
\(\displaystyle (6+2m)^2-4(m^2+1)(5)=0\)
\(\displaystyle 36+24m+4m^2-20m^2-20=0\)
\(\displaystyle -16m^2+24m+16=0\)
\(\displaystyle 2m^2-3m-2=0\)
\(\displaystyle 2m^2+m-4m-2=0\)
m(2m + 1) - 2(2m + 1) = 0
(m - 2)(2m + 1) = 0
m - 2 = 0 or 2m + 1 = 0
m = 2 or 2m = -1
After I rechecked the question, I did some typos and the option D was actually 2, so that is the answer.