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1. Can you explain/go through how to do part b) please?

Thanks.

2. Consider the relationship between a radius and tangent
3. (Original post by TenOfThem)
Consider the relationship between a radius and tangent
Yes I did that Please elaborate.
4. Well that gives you the gradient

Then a bit of Pythagoras will find BC
5. a) (x-8)^2+(y-4)^2=25
b)i)m(FB)=4/3 and equation of tangent BC is 3x+4y=65
ii)37.74cm

You don't need pythagoras. Unless by pythagoras you mean equation to calculate a line so pythagoras twice, but I doubt OP knows that that is pythagoras (no offense OP)
6. (Original post by GreenLantern1)
a) (x-8)^2+(y-4)^2=25
b)i)m(FB)=4/3 and equation of tangent BC is 3x+4y=65
ii)37.74cm

You don't need pythagoras. Unless by pythagoras you mean equation to calculate a line so pythagoras twice, but I doubt OP knows that that is pythagoras (no offense OP)
Just realised I made an error. Last bit is 28.4
7. (Original post by GreenLantern1)
a) (x-8)^2+(y-4)^2=25
b)i)m(FB)=4/3 and equation of tangent BC is 3x+4y=65
ii)37.74cm

You don't need pythagoras. Unless by pythagoras you mean equation to calculate a line so pythagoras twice, but I doubt OP knows that that is pythagoras (no offense OP)

(Original post by TenOfThem)
Well that gives you the gradient

Then a bit of Pythagoras will find BC
I don't know why the gradient of FB = 4/3 ???

Yes I do understand the Pythagoras bit
8. (Original post by blueray)
I don't know why the gradient of FB = 4/3 ???

Yes I do understand the Pythagoras bit
You know F and B so the gradient is (change in y)/(change in x)
9. (Original post by GreenLantern1)
You don't need pythagoras.
I am wondering how you think you can find the length of BC without Pythagoras
10. (Original post by TenOfThem)
I am wondering how you think you can find the length of BC without Pythagoras
Don't use selective treatment to quote me. You know full well how I calculated it.

What I meant was you don't need to do a^2+b^2=c^2 in that format. You can just use the equation:

(x1-x2)^2+(y1-y2)^2 (this is all square rooted). And that I said is derived from pythagoras!
11. (Original post by TenOfThem)
You know F and B so the gradient is (change in y)/(change in x)
(8-4)/(11-8)=4/3

I would rather not confuse the OP by getting into a nonsense argument
13. (Original post by GreenLantern1)
(8-4)/(11-8)=4/3

(Original post by TenOfThem)

I would rather not confuse the OP by getting into a nonsense argument
Ok so I understand how to get FB but

why is grad of tangent at B = -3/4
14. Because radius and tangent meet at a right angle
15. (Original post by TenOfThem)
Because radius and tangent meet at a right angle
Yes I know that? But why would it become a negative reciprocal then??
16. (Original post by blueray)
Yes I know that? But why would it become a negative reciprocal then??
Because they meet at a right angle

And that is the rule for perpendicular gradients
17. (Original post by GreenLantern1)
Just realised I made an error. Last bit is 28.4
I got 28.2. But that must be due to me using incredibly long decimals
18. (Original post by CharlieBoardman)
I got 28.2. But that must be due to me using incredibly long decimals
No if you were to use incredibly long deciamls you would have like 28.43333333. You have just done something worng and it seems fudged your results to get near the given.

19. (Original post by blueray)
Yes I know that? But why would it become a negative reciprocal then??
You should know that if gradient A is perpendicular to gradient B, then gradient B works out with that coreelating thing a s a product to -1.
20. (Original post by GreenLantern1)
No if you were to use incredibly long deciamls you would have like 28.43333333. You have just done something worng and it seems fudged your results to get near the given.

Ah, I messed up on the equation of the tangent at B! -.- I was 0.75 out on the intercept. I rushed through it, which altered my results. Yes, 28.403333... Is correct

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