Posted by xavier on Thursday, January 31, 2013 at 8:39am.
let the centre be (a,b)
but it must lie on x+2y=0, so a+2b=0
a = -2b
we can call our centre (-2b, b)
its distance to 3x + 4y - 5 must be 1
| 3(-2b) + 4b - 5 }/√(3^2+4^2) = 1
case 1:
(-6b+4b-5)/√25 = 1
-2b - 5 = 5
b = -5
a= 10 ----> equation: (x-10)^2 + (y+5)^2 = 1
case 2:
6b - 4b + 5 = 5
2b = 0
b=0
a=0 ----------> x^2 + y^2= 1
make a sketch to see if both are possible
What do you think?
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