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2007 Past Paper Question - HSN forum

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2007 Past Paper Question


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#1 Philip

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Posted 18 May 2009 - 12:09 PM

Just been doin a few past paper questions and got a little stuck... blink.gif

2007 Paper 1 Question 5

I get the centre of the circle. But i don't know what to do from there!

Would appreciate some help, thanks wink.gif
Yeah!

#2 christie

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Posted 18 May 2009 - 12:40 PM

You need to find the centre B first, (7,8).

Then find the diameter of the first big circle, 12.

That means the radius of each small circle is 2.

B is (7,8), so D must be (15,8) because there is four radii between B and D.

Use the equation..

(x-a)2 + (y-b)2 = r2

(x-15)2 + (y-8)2 = 22

#3 Ailso

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Posted 18 May 2009 - 12:53 PM

This is how i got it...not sure its the correct way though, but i got the answer. (Sorry if its messy dont know how to do the shortcut things on this site! Also, i dont know where the marks will be allocaed either.)

1) Work out point B to be (7,8)

2) Radius of the large circle = square root of ((g squared) +(f squared) - c)

i.e,square root of ((7squared)+(8squared)-77) = square root of 36 = 6

therefore radius of large circle = 6

3) Between the point B and the outer of the circle there are three sets of radii this means that the radii of the smaller
circles is the radius of the large circle divided by three, i.e 6 divided by 3=2, this is the length of each of the small radii
because the question tells us they are congruent.

4) The y-coordinate is the same for each of the centres cause they are parallel to the x-axis. So for D y=8

5) To get the x-coordinate for D use the x-coordinate from B. So it would be x-coordinate + radius of large circle + radius
of small circle, i.e 7+6+2=15.

6) The point D has coordinates (15,8)

7) Put these numbers into the equation of a circle ((x-a)squared + (y-b)squared = radius squared)to give us:

(x-15)squared + (y-8)squared = 2squared

Hope this helps, if you need anymore explanation, just ask! cool.gif
My personal hobbies are reading, listening to music, and silence - Edith Sitwell

#4 Philip

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Posted 20 May 2009 - 10:38 AM

QUOTE (Ailso @ May 18 2009, 01:53 PM) <{POST_SNAPBACK}>
This is how i got it...not sure its the correct way though, but i got the answer. (Sorry if its messy dont know how to do the shortcut things on this site! Also, i dont know where the marks will be allocaed either.)

1) Work out point B to be (7,8)

2) Radius of the large circle = square root of ((g squared) +(f squared) - c)

i.e,square root of ((7squared)+(8squared)-77) = square root of 36 = 6

therefore radius of large circle = 6

3) Between the point B and the outer of the circle there are three sets of radii this means that the radii of the smaller
circles is the radius of the large circle divided by three, i.e 6 divided by 3=2, this is the length of each of the small radii
because the question tells us they are congruent.

4) The y-coordinate is the same for each of the centres cause they are parallel to the x-axis. So for D y=8

5) To get the x-coordinate for D use the x-coordinate from B. So it would be x-coordinate + radius of large circle + radius
of small circle, i.e 7+6+2=15.

6) The point D has coordinates (15,8)

7) Put these numbers into the equation of a circle ((x-a)squared + (y-b)squared = radius squared)to give us:

(x-15)squared + (y-8)squared = 2squared

Hope this helps, if you need anymore explanation, just ask! cool.gif


Thanks! wink.gif

Yeah!





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