Question

Show that the straight line with equation 2x+y-8 = 0 is a tangent to a circle with equation x^2 + y^2 + 4x - 4y - 12 = 0. What is the equation of the radius drawn to the point of contact?

Answer 1

2x + y - 8 = 0
2x - 2x + y - 8 + 8 = 0 - 2x + 8
y = -2x + 8

x² + (-2x + 8)² + 4x - 4(-2x + 8) - 12 = 0
x² + (-2x + 8)(-2x + 8) + 4x - 4(-2x) - 4(8) - 12 = 0
x² + (4x² - 16x - 16x + 64) + 8x - 32 - 12 = 0
x² + 4x² - 32x + 64 + 8x - 32 - 12 = 0
x² + 4x² - 32x + 8x + 64 - 32 - 12 = 0
5x² - 24x - 20 = 0
x = -(-24) +/- √((-24)² - 4(5)(-20))
2(5)
x = 24 +/- √(196 + 400)
10
x = 24 +/- √(596)
10
x = 24 +/- 24.413112314674
10
x = 2.4 +/- 2.4413112314674
x = 2.4 + 2.4413112314674 x = 2.4 - 2.4413112314674
x = 4.8413112314674 x = -0.043112314674

y = -2x + 8 y = -2x + 8
y = -2(4.8413112314674) + 8 y = -2(-0.043112314674) + 8
y = -9.6826224629348 + 8 y = 0.086224629348 + 8
y = -1.6836224629348 y = 8.086224629348
(x, y) = (4.8413112314674, -1.683622462938) (x, y) = (-0.043112314674, 8.086224629348)

Answer 2

`2x+y-8=0\\ y=-2x+8`

Now put it into the equation of a circle.


`x^2+(-2x+8)^2+4x-4(-2x+8)-12=0\\ x^2+4x^2-32x+64+4x+8x-32-12=0\\ 5x^2-20x+20=0\\ 5(x^2-4x+4)=0\\ 5(x-2)^2=0\\ (x-2)^2=0\\ x-2=0\\ x=2\\\\ y=-2\cdot2+8=-4+8=4\\\\(x,y)=(2,4)`

The solution is only one point which means the line is tangent to the circle.