Maths

Let B and D be the intersection of C and x-axis, C and T respectively.
Coordinates of O = (1, -2)
So tanOAB = 2/(1 + 6) = 2/7
Note that angle OAB = angle OAD
So tanDAB = tan2OAB
= 2tanOAB/(1 - tan^2OAB)
= 2(2/7)/(1 - 4/49)
= 28/45
So slope of T = -28/45
Equation of T is
y = (-28/45)(x+6)
28x + 45y + 168 = 0

2010-11-13 12:05:09 補充：
tanDAB = 28/45
Slope = tan(pi - DAB) = -tanDAB
= -28/45

2010-11-13 12:13:29 補充：
Alternatively,
Let the slope of T be m, then equation of T is
y = m(x + 6)
Substitute in the equation of C:
x^2 + m^2(x + 6)^2 - 2x + 4m(x + 6) + 1 = 0
(1 + m^2)x^2 + (12m^2 + 4m - 2)x + 36m^2 + 24m + 1 = 0

2010-11-13 12:13:34 補充：
For tangency, delta = 0
(12m^2 + 4m - 2)^2 - 4(36m^2 + 24m + 1)(1 + m^2) = 0
-180m^2 - 112m = 0
m = 0 or -28/45
So the equation of T is
y = (-28/45)(x + 6)
28x + 45y + 168 = 0

I'm sorry I have made a mistake
C: x^2 + y^2 - 2x + 4y + 1 = 0

Part of the equation of the circle is missed.