Posted by tembo on Wednesday, March 14, 2012 at 10:49am.
x^2+2x+1 + y^2-8y+16=1+16-5
(x+1)^1 + (y-4)^2=12
center, -1,4
radius sqrt12
complete the square ....
x^2 + 2x + .... + y^2 - 8y + ... = -5
x^2 + 2x + 1 + y^2 - 8y + 16 = -5+1+16
(x+1)^2 + (y-4)^2 = 12
centre is (-1,4) , radius is √12 or 2√3
if x-3y=8 is a tangent, then the distance from (-1,4) to the line should be √12
distance from (-1,4) to x-3y-8=0 is
|-1 - 3(4) - 8|/√(1+9)
= 21/√10 ≠ √12
so, no, it is not a tangent.
other way, find intersection....
x = 3y+8 , sub into circle
(3y+8)^2 + y^2 + 2(3y+8) -8y+5=0
9y^2 + 48y + 64 + y^2 + 6y + 24 - 8y + 5 = 0
10y^2 +46y +93=0
to be tangent, this quadratic should have one solution, thus it should be a perfect square.
But it is not, so .... no tangent
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