請問有關數學

I will give two methods here because I am not sure if you know differentiation. If you don't know differentiation, just skip Method 1 and go to Method 2.

Method 1: (Differentiation)
x² + y² + 4x + 2y - 20 = 0
Differentiate both sides with respect to x,
2x + 2y dy/dx + 4 + 2 dy/dx = 0
x + y dy/dx + 2 + dy/dx = 0
dy/dx = -(x+2) / (y+1)

Put x = 1, y = 3, we can get the slope of tangent line (the line touching) at A.
dy/dx|(x=1, y=3) = -(1+2) / (3+1) = -3/4

So the equation of the tangent is :
-3/4 = (y-3)/(x-1)
-3(x-1) = 4(y-3)
3x+4y-15 = 0


Method 2: (Solving quadratic equation)
Let the equation of the tangent be y = mx + c.
Put x = 1 and y = 3,
3 = m(1) + c
c = 3-m
So the equation is y = mx + (3-m)

Put it into the equation of the circle.
x² + [mx + (3-m)]² + 4x + 2[mx + (3-m)] - 20 = 0
x² + m²x² + 2(3-m)mx + (3-m)² + 4x + 2mx + 2(3-m) - 20 = 0
(1+m²)x² + (-2m²+6m+2m+4)x + (3-m)² + 2(3-m) - 20 = 0
(1+m²)x² - 2(m²-4m-2)x + (m²-8m-5) = 0
Since the line touches the circle, Δ=0
4(m²-4m-2)² - 4(1+m²)(m²-8m-5) = 0
4(m^4 + 16m² + 4 - 8m³ - 4m² + 16 m) - 4(m²-8m-5+m^4 -8m³ -5m²) = 0
m^4 + 16m² + 4 - 8m³ - 4m² + 16 m - (m²-8m-5+m^4 -8m³ -5m²) = 0
16m² + 24m + 9 = 0
(4m+3)² = 0
m = -3/4 (Repeated root)

So the equation of the tangent is
y = (-3/4)x + [3-(-3/4)]
y = -3/4 x + 15/4
4y = -3x + 15
3x + 4y - 15 = 0


Hope it helps! ^^