C1: x^2 + y^2 + 2gx +2fy +c=0

C2: x^2 + y^2 + 2g'x +2f'y +c'=0

p and q are the centers of C1 and C2 respectively, show that 2gg'+ 2ff' = c+ c'

please is someone could help that would be amazing thank you!

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- Feb 19th 2008, 02:02 PMIlusaMaths exam tomorrow need help with a circle proof please
C1: x^2 + y^2 + 2gx +2fy +c=0

C2: x^2 + y^2 + 2g'x +2f'y +c'=0

p and q are the centers of C1 and C2 respectively, show that 2gg'+ 2ff' = c+ c'

please is someone could help that would be amazing thank you! - Feb 19th 2008, 02:15 PMPlato
- Feb 19th 2008, 02:23 PMIlusa
Oh my goodness I'm so sorry.

there is meant to be more

C1: x^2 + y^2 + 2gx +2fy +c=0

C2: x^2 + y^2 + 2g'x +2f'y +c'=0

are two circles

p and q are the centers of C1 and C2 respectively, and t is the point of intersection of the two circles.

given that pt is perpendicular to qt, show that show that 2gg'+ 2ff' = c+ c' - Feb 19th 2008, 02:39 PMPlato
- Feb 19th 2008, 02:50 PMIlusa
sorry I dont' get this bit

the points p and q are not given but looking at the diagram the radii of the two circles are equal. - Feb 19th 2008, 03:11 PMPlato
- Feb 19th 2008, 03:15 PMIlusa
oh thank you, well I wrote that distance as (r^2+r'^2)^1/2

as distance pt=radius of C1

and distance qt=raius of C2

I drew a right angled triangle and used pytagoras' therom to get distance pq

but I still can't prove it - Feb 19th 2008, 03:54 PMPlato
- Feb 19th 2008, 04:05 PMIlusa
thank you so much, you're so kind and intelligent! =)

sorry to bother you but where do you get (g'-g)^2 + (f'-f)^2 from? - Feb 19th 2008, 04:08 PMIlusa
oh sorry it was the distance formula squared sorry! Thank you so much again!

- Feb 19th 2008, 04:12 PMPlato
- Feb 19th 2008, 04:14 PMIlusa
yeah I do thanks, I understand it fully now thank you again!