Question 989033
Centre of {{{x^2 + y^2 =r^2}}} is (0,0) and radius r. 
Since y=mx+c (or say mx-y+c=0) is tangent to this circle, distance from centre(0,0) to this line is equal to radius
So,
{{{(m*0-0+c)/sqrt(m^2+(-1)^2) = r}}}
=> {{{c/sqrt(m^2+1) = r}}}
=> {{{c = r*sqrt(m^2+1)}}} --------------part (i)

Let y=mx+c be tangents to circle x^2 + y^2 = 4
Since, it passes through (0,6) and (0,-6)
So,
6=0+c  and -6 = 0+c
=> c=6 or -6
Now, using part(i) proof,
{{{6=2*sqrt(m^2+1)}}} or {{{-6=2*sqrt(m^2+1)}}}
=> {{{m=2sqrt(2)}}} or {{{m=-2sqrt(2)}}} (from both equations-same values of m)
So equation of tangents are
{{{y=2sqrt(2)x+6}}} and {{{y=2sqrt(2)x-6}}} and {{{y=-2sqrt(2)x+6}}} and {{{y=-2sqrt(2)x-6}}}

Since there are two external points, hence two tangents can be drawn from each point. So, we have got 4 equations of tangents.