tangent lines of a circle

find tangent lines through a point (a,b) out of a circle x^2+y^2=r^2.

 

tangent lines : y = m(x-a)+b ⇔ mx - y - am + b = 0, where m represents the gradient of the the tangent lines.

 

the distance between the point (0,0) which is the center of the circle and the tangent lines is

 

|-am+b|

--------------

root(m^2+1)

= r


⇔  (a^2-r^2)m^2-2abm+b^2-r^2=0

⇔  m = 

ab ± root(r^2(a^2+b^2-r^2))

-------------------------------------

a^2-r^2

 

therefore, the tangent lines are

 

y = 

ab ± root(r^2(a^2+b^2-r^2))

------------------------------------- (x-a) + b

a^2-r^2

 


<question>

find the tangent lines through a point (-4,1) out of a circle x^2+y^2-4x-2y-4=0.

 

<answer>

x^2+y^2-4x-2y-4=0 ⇔ (x-2)^2+(y-1)^2=9

 

moving the center of the circle (2,1) to (0,0),

the point (-4,1) to(-6,0).

 

therefore

 m = 

0 ± root(9(36+0-9))

-------------------------------

36-9

= ±1/root(3) 

 

therefore, by m=±1/root(3) and (-4,1) ,

 

y = ±1/root(3)(x + 4) + 1

 

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