maths
posted by tembo .
Given that the equation of the circle is x^2+y^2+2x8y+5=0
(i)find the centre and radius of the circle
(ii)Determine whether the line x3y=8 is tangent to the given circle

x^2+2x+1 + y^28y+16=1+165
(x+1)^1 + (y4)^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 + (y4)^2 = 12
centre is (1,4) , radius is √12 or 2√3
if x3y=8 is a tangent, then the distance from (1,4) to the line should be √12
distance from (1,4) to x3y8=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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