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Point of contact of two circles Help needed

Tracy

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

Steve

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.

Tracy

Posted 26 January 2006 - 07:01 PM

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

Thanks x x x

Steve

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


Tracy

Posted 26 January 2006 - 07:26 PM

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

Steve

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).

Tracy

Posted 26 January 2006 - 07:33 PM

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


Steve

Posted 26 January 2006 - 07:37 PM

Yes, that's what I get. smile.gif

duncad

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).


ChrisEss

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 :\

Nathan

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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