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

**0**

# Point of contact of two circles

Started by Tracy, Jan 26 2006 06:45 PM

10 replies to this topic

### #1

Posted 26 January 2006 - 06:45 PM

### #2

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.

### #3

Posted 26 January 2006 - 07:01 PM

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

Thanks x x x

Thanks x x x

### #4

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.

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

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

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

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

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

### #5

Posted 26 January 2006 - 07:26 PM

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

### #6

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).### #7

Posted 26 January 2006 - 07:33 PM

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

### #8

Posted 26 January 2006 - 07:37 PM

Yes, that's what I get.

### #9

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).

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

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

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 :\

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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