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# Circles watch

1. Hi,
Question:
Circle has centre (5,8) express equation in form:

(x-a)^2 + (b-y)^2 = k

The answer is: (x-5)^2 + (b-8)^2 = 25

Could anyone explain how it equals 25? I got everything right apart from I got that it equals -89.
Thanks very much
2. (Original post by daviem)
Hi,
Question:
Circle has centre (5,8) express equation in form:

(x-a)^2 + (b-y)^2 = k

The answer is: (x-5)^2 + (b-8)^2 = 25

Could anyone explain how it equals 25? I got everything right apart from I got that it equals -89.
Thanks very much
There must be more information. A centre is not enough to define a unique circle.
3. Was there radius given? If the answer is "(x-5)^2 + (b-8)^2 = 25" then the radius could have been given as 5.
4. There was nothing else other than a picture but as far as I can see theres no extra info on it.
5. mind showing us the picture? or the whole question
6. That's the picture, sorry I don't know how else to upload it because its on my laptop
7. If you want to look at the exam paaper its AQA pure core 1, question 6, may 2012
8. In the photo given, the circle touches the y axis, and has y-coordinate of 5. Therefore the radius of the circle is 5!
9. (Original post by daviem)
There was nothing else other than a picture but as far as I can see theres no extra info on it.
The circle touches the y-axis : key peice of information that you didn't tell us.

Can you see what the radius of the circle must be?
10. ah, well you see it touches the y axis. therefore the radius must be from the center to the y axis which is equal to its x co-ordinate, which equals 5. and the first part of the equation always equals the radius squared
11. Circle equations are (x-a)^2 + (y-b)^2 = r^2
for a circle with radius "r" and centre (a,b)
12. It touches the Y axis which means that the radius has to be equal to the distance the centre is from it. In this case that means the radius is 5.

(x-5)^2 + (x-8)^2 = 25
13. Ah, right yeah, sorry, don't know why id didn't see that, thanks very much, I really appreciate all your help

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