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# C2 coordinate geometry question watch

1. Hi all.

I was working through the mixed exercise for the C2 edexcel course and was wondering if I could get some help on the following question:

Q14) The centres of the circles (x-8)^2+(y-8)^2=117
and
(x+1)^2 + (y-3)^2=106, are P and Q respectively.

a) show that P lies on (x+1)^2 + (y-3)^2=106
b) find the length of PQ

Cheers
2. (Original post by Meha)
Hi all.

I was working through the mixed exercise for the C2 edexcel course and was wondering if I could get some help on the following question:

Q14) The centres of the circles (x-8)^2+(y-8)^2=117
and
(x+1)^2 + (y-3)^2=106, are P and Q respectively.

a) show that P lies on (x+1)^2 + (y-3)^2=106
b) find the length of PQ

Cheers
For part a) you need to first work out the centre of the circle P which has eqn:
(x-8)^2+(y-8)^2=117

Once you do that plug the x and y co-ordinates into the eqn :
(x+1)^2 + (y-3)^2=106

if the answer is =106 then you know that point LIES on the circle. had it been <106 it would lie inside.

for B you need to first work out the centre of the 2 circles. Then use pythagoras's theorem to work out the distance with this formula:
3. For (a) plug in the coordinates of P into that equation. If the LHS = the RHS then it must lie on that circle.

For (b), notice that Q is the centre of the circle that P lies on, and so the length of PQ must be the length of the radius of the circle with centre Q.
4. (Original post by nuodai)
For (a) plug in the coordinates of P into that equation. If the LHS = the RHS then it must lie on that circle.

For (b), notice that Q is the centre of the circle that P lies on, and so the length of PQ must be the length of the radius of the circle with centre Q.

Lol just noticed this :P
5. You know the x and y coords of P, so plug those into the equation and see if LHS = RHS.

Can you remember the equation to find the distance between two points? (Think pythagoras)
6. a) show that P lies on (x+1)^2 + (y-3)^2=106
b) find the length of PQ

-------------
a) Substitute the X/Y values of point P into the equation of the circle , and it should equal to 106 . If it does, it lies on it.

b) Use the distance formula from the centre to the point.
7. Guyz, you definitely don't need to use Pythagoras or distance formulae to find the distance between P and Q in part (b).
8. Ah yes, if it lies on the circle the distance is simply the radius.

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