# Circles C2Watch

#1
Hey -

Thanks
0
9 years ago
#2
Can you post it any other way that pic isn't appearing for me?
0
9 years ago
#3
If you could draw a diagram of the circle with a tangent and the centre of the circle marked on, you'd get a right-angled triangle with the centre (I'll call it C) to B being at right angles to AB.

So you could work out the length of AC (you know their coordinates) and as you know the length of the radius BC, you can use pythagoras to find AB.

Does that kind-of match the mark scheme? I've tried to attach a diagram but I don't know if it worked!
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#4
(Original post by fishpie57)
If you could draw a diagram of the circle with a tangent and the centre of the circle marked on, you'd get a right-angled triangle with the centre (I'll call it C) to B being at right angles to AB.

So you could work out the length of AC (you know their coordinates) and as you know the length of the radius BC, you can use pythagoras to find AB.

Does that kind-of match the mark scheme? I've tried to attach a diagram but I don't know if it worked!
Sorry, how do you know you would get a right angled triangle?
0
9 years ago
#5
(Original post by Darkest Knight)
Sorry, how do you know you would get a right angled triangle?
A tangent meets the circle at 90degrees.

Plus tangents are perpendicular to the normal and so that also tells you 90degrees.
0
9 years ago
#6
isn't this C4?? you have to implicitly differentate 1st then stick coodindates in
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9 years ago
#7
(Original post by Darkest Knight)
Hey -

Thanks
Well, youv'e been given x and y, so sub that in to find k. You've done, that, so part c is with normals and tangents!
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9 years ago
#8
(Original post by Darkest Knight)
Sorry, how do you know you would get a right angled triangle?
GCSE Circle Theorems - a tangent meets a radius at 90 degrees. BC is a radius. AB is a tangent. So it's therefore 90 degrees. Thus you can use pythagoras!
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