Given
#p = (x,y)#
#p_1 = (x_1,y_1)=(-1,4)#
#p_2 =(x_2,y_2)= (6,3)#
#vec v = (m,1)#
#r = 5#
with the line #L# and circle #C#
#L->p = p_1 + lambda vec v#
#C->norm(p-p_2) = r#
we have the intersection #L nn C# is given by the solution of
#norm(p_1-p_2+lambda vec v) = r# or squaring both sides
#norm(p_1-p_2)^2+2lambda << p_1-p_2, vec v >> + lambda^2 norm(vec v)^2 = r^2#
solving for lambda
#lambda = 1/(2 norm(vec v)^2) (-2 << p_1-p_2, vec v >> pm sqrt(4<< p_1-p_2, vec v >> ^2-4norm(vec v)^2(norm(p_1-p_2)^2-r^2)))#
If #C# and #L# are tangent then necessarily
#4<< p_1-p_2, vec v >> ^2-4norm(vec v)^2(norm(p_1-p_2)^2-r^2)=0#
or
#((x_1-x_2)m+(y_1-y_2))^2-(m^2+1)((x_1-x_2)^2+(y_1-y_2)^2-r^2) = 0#
now solving for #m#
#m = (pmr sqrt[(x_1 - x_2)^2 + (y_1 - y_2)^2-r^2) - (x_1 - x_2) (y_1 -
y_2))/((r + y_1 - y_2) (r - y_1 + y_2))#
solving for the given data we obtain
#m = {-3/4, 4/3}#
