Thursday

October 30, 2014

October 30, 2014

Posted by **xavier** on Thursday, January 31, 2013 at 8:39am.

- Analytic Geometry -
**Reiny**, Thursday, January 31, 2013 at 9:30amlet 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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