find the equation of tangent

As the tangent passes through (-19, 9), so, let the equation of the tangent bey - 9 = m(x + 19)==> y = mx + 19m + 9 ⋯⋯⋯ (from graph, m > 0)Substitute into the circle equation, we get,x² + y² + 28x - 20y = -288==> x² + (mx + 19m + 9)² + 28x - 20(mx + 19m + 9) + 288 = 0==> x² + m²x² + 38m²x + 18mx + 361m² + 342m + 81 + 28x - 20mx - 380m - 180 + 288 = 0==> (m² + 1)x² + (38m² - 2m + 28)x + (361m² - 38m + 189) = 0As it is a tangent, so the discriminant of this equation is 0, that is,(38m² - 2m + 28)² - 4(m² + 1)(361m² - 38m + 189) = 0==> (19m² - m + 14)² = (m² + 1)(361m² - 38m + 189)==> 361m⁴ - 38m³ + 533m² - 28m + 196 = 361m⁴ - 38m³ + 550m² - 38m + 189==> 17m² - 10m - 7 = 0==> (m - 1)(17m + 7) = 0==> m = 1 or m = -7/17 (rej. as m > 0)
So the equation of tangent is :y = x + 19 + 9==> x - y + 28 = 0

Thank you very much YA HOO ! 知識+管理員 !!!

Other methods：
http://postimg.org/image/9c6vxqdt7/full/

2014-11-09 17:36:12 補充：
M4：
http://postimg.org/image/i24w1bt63/

2014-11-09 18:03:04 補充：
M5：
http://postimg.org/image/abeus1kjd/

2014-11-09 21:48:56 補充：
You are welcome ~~

To skip the differentiation part :
let the tangent touches the circle at (a,b).
Center of circle is (-14, 10).
Slope of line connecting (a,b) and center = (b - 10)/(a + 14).
Slope of tangent = (b - 9)/(a + 19)
So (b - 10)/(a + 14) x (b - 9)/a + 19) = - 1
Simplifying we get 5a + b + 68 = 0......................(1)
Since (a,b) is on the circle, so
a^2 + b^2 + 28a - 20b = - 888 ................(2)
This 2 equations are the same as previously derived, so the remaining part is the same, so will not repeat here.

可以用 Δ = 0，有勞 HK~ 講解一下。

2014-11-09 14:19:18 補充：
y - 9 = m(x + 19)

代入 circle equation

令 Δ = 0

計出 m

成功！