Tangents to a circle

The Qs are solved as follows:

圖片參考：http://i238.photobucket.com/albums/ff245/chocolate328154/Maths259.jpg?t=1212491530

Let tangent to the circle and passes through the point (-1,-2) be y + 2 = m(x + 1), that is mx - y +(m-2) = 0. Centre of the circle is (-4,-3), radius of the circle is sqrt5. So distance from (-4,-3) to the tangent equals to sqrt5. That is
[m(-4) -(-3) +(m-2)]/sqrt[m^2 +(-1)^2] = +/-sqrt5
(-4m +3 +m -2)/sqrt[m^2 1] = /-sqrt5
(-3m + 1)^2/(m^2 +1) = 5
(9m^2 + 1 -6m)/(m^2 + 1) = 5
9m^2 + 1 -6m = 5m^2 + 5
4m^2 -6m -4 = 0
2m^2 -3m -2 = 0
(2m + 1)(m-2) = 0
m = 2 or -1/2
Therefore, the 2 tangents are:
2x -y = 0
-x/2 -y -5/2 = 0 or x + 2y + 5 = 0
Same method can be applied to the other question.