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Point of contact of two circles - HSN forum

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Point of contact of two circles


10 replies to this topic

#1 Tracy

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Posted 26 January 2006 - 06:45 PM

Got a question i am stuck with:

Circle A has equation x2 + y2 - 10x - 12y + 36 =0. Circle B has equation x2 + y2 + 8x + 12y -48 = 0.

(a) Find the circle and radius of each circle
(b) Show that the circles touch externally
( c) Find the point of coordinates of the point of contact of the circles


I can do (a) and (b) BUT © is just impossible. Thought it would be better to put the whole question up.

Help Please, thanks trace x x

#2 Steve

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Posted 26 January 2006 - 06:56 PM

QUOTE(Tracy @ Jan 26 2006, 07:45 PM)
( c) Find the point of  coordinates of the point of contact of the circles

I'm pretty sure you cover this in the Vectors outcome of Unit 3, but I can try explaining it to you if you like.

Edit: In fact, I don't think you know enough yet for me to explain it. I could be wrong but I don't see another way of doing it. You should cover it in Unit 3 though.
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#3 Tracy

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Posted 26 January 2006 - 07:01 PM

Can you give it a bash please im resitting it this year

Thanks x x x

#4 Steve

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Posted 26 January 2006 - 07:19 PM

Ok, have a look at the attachment for a picture of what's going on. You could just sketch this roughly.

Remember from Vectors that you can find a point which divides a line in a ratio. That's all that's happening here.



\begin{align*}
\frac{\mathrm{BP}}{\mathrm{PA}} &= \frac{10}{5} \\
\overrightarrow{\mathrm{BP}}&=2\overrightarrow{\mathrm{PA}}
\end{align*}

Does this ring any bells? (You might have used the Section Formula also).

I'll go further if you still don't get it smile.gif.

Attached Thumbnails

  • Attached Image: circles.jpeg

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#5 Tracy

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Posted 26 January 2006 - 07:26 PM

Will that not just give you one of the needed points for the co-ordinate?

#6 Steve

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Posted 26 January 2006 - 07:28 PM

If you expand those out and use position vectors, then you can rearrage to find p, which is the position vector of the point of contact (there is only one since the circles touch externally).
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#7 Tracy

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Posted 26 January 2006 - 07:33 PM

I get the answer (2,2) do u agree


#8 Steve

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Posted 26 January 2006 - 07:37 PM

Yes, that's what I get. smile.gif
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#9 duncad

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Posted 27 January 2006 - 07:28 AM

If this question has been raised in units 1 or 2 then you are not expected to use vectors.

Centre A = (5,6) Radius A = 5
Centre B= (-4,-6) Radius B = 10

Draw a sketch (see Steves) and simply use similar triangles, one with hyp = 15 and the other hyp = 10. It is then easy to work out the p.o.c. (2,2).


#10 ChrisEss

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Posted 28 January 2006 - 12:50 AM

or you coul djus substitute both equations for an x value and then sub the x value back into one of the equations for a y value :\

#11 Nathan

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Posted 28 January 2006 - 01:04 AM

QUOTE(ChrisEss @ Jan 28 2006, 12:50 AM)
or you coul djus substitute both equations for an x value and then sub the x value back into one of the equations for a y value :\

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i dont think that would work but in saying that it is 1am!!....would it work?

anyway, i would use duncads method





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