Posted by **tembo** on Wednesday, March 14, 2012 at 10:49am.

Given that the equation of the circle is x^2+y^2+2x-8y+5=0

(i)find the centre and radius of the circle

(ii)Determine whether the line x-3y=8 is tangent to the given circle

- maths -
**bobpursley**, Wednesday, March 14, 2012 at 10:51am
x^2+2x+1 + y^2-8y+16=1+16-5

(x+1)^1 + (y-4)^2=12

center, -1,4

radius sqrt12

- maths -
**Reiny**, Wednesday, March 14, 2012 at 10:59am
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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