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# Is the radius of this circle equation not 2? watch

1. (x-3/4)^2 + (y+1/4)^2 = 1

The book says 4, I keep getting 2.

?
2. Surely it's 1?
3. Err, I could be misremembering something, but isn't the radius of that circle just 1?...
4. Is that or ?

I'd guess it's the former from your question, in which case remember that if you take the 4 out of the brackets, you need to square it.
5. Do you mean or . If it's the former, the answer's clearly 1. If it's the latter, remember that , not r.
6. (Original post by BJack)
Is that or ?

I'd guess it's the former from your question, in which case remember that if you take the 4 out of the brackets, you need to square it.
No, it's the first
The book says 4...the centre is (3,-1)
7. (Original post by Get Cape.Wear Cape.Fly.)
No, it's the first
The book says 4...the centre is (3,-1)
The book is right, then. Multiple both sides by 4^2 to get it into the form (x-a)^2 + (y-b)^2 = r^2, and you find r^2 = 16...
8. (Original post by Get Cape.Wear Cape.Fly.)
No, it's the first
The book says 4...the centre is (3,-1)
((x-3)/4)^2 + ((y+1)/4)^2 = 1
to make it more clear, we can say it is actually:
(a(x-3))^2 + (a(x+1))^2 = 1, where a=(1/4)
whe you take something out of brackets being squared, you square it
so:
a^2(x-3)^2 + a^2(x+1)^2=1
a^2=1/16
so multiplying both side by 16, will get rid of the a^2
(x-3)^2 + (x+1)^2=16

as (x+a)^2 + (x+b)^2=r^2

r^2=16
r=4

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