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, 01:02 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, 01:15 PMPlato
- Feb 19th 2008, 01:23 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, 01:39 PMPlato
Let $\displaystyle \left( {r_p } \right)\,\& \,\left( {r_q } \right)$ be the radii of the circles centered at

*p & q*respectively.

By completing squares we get $\displaystyle \left( {r_p } \right)^2 = g^2 + f^2 - c\;\,\& \,\,\left( {r_q } \right)^2 = \left( {g'} \right)^2 + \left( {f'} \right)^2 - c'$.

From the given $\displaystyle \left[ {l\left( {\overline {pg} } \right)} \right]^2 = \left( {r_p } \right)^2 + \left( {r_q } \right)^2 $

What points are P & q?

Can you finish? - Feb 19th 2008, 01:50 PMIlusa
sorry I dont' get this bit

$\displaystyle

\left[ {l\left( {\overline {pg} } \right)} \right]^2

$

the points p and q are not given but looking at the diagram the radii of the two circles are equal. - Feb 19th 2008, 02:11 PMPlato
- Feb 19th 2008, 02:15 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, 02:54 PMPlato
$\displaystyle \left[ {l\left( {\overline {pg} } \right)} \right]^2 = \left( {g' - g} \right)^2 + \left( {f' - f} \right)^2 = \left( {g^2 + f^2 - c} \right) + \left( {\left( {g'} \right)^2 + \left( {f'} \right)^2 - c'} \right)$

- Feb 19th 2008, 03:05 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, 03:08 PMIlusa
oh sorry it was the distance formula squared sorry! Thank you so much again!

- Feb 19th 2008, 03:12 PMPlato
- Feb 19th 2008, 03:14 PMIlusa
yeah I do thanks, I understand it fully now thank you again!