Analytic Geometry
posted by xavier .
Find the equation of the circle, with radius 1, tangent to the line 3x+4y=5 and having its center on the line x+2y=0

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+4b5)/√25 = 1
2b  5 = 5
b = 5
a= 10 > equation: (x10)^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?