Maths exam tomorrow need help with a circle proof please

**Ilusa** · February 19th 2008, 01:02 PM

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!

**Plato** · February 19th 2008, 01:15 PM

Originally Posted by Ilusa

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'

As written, I do not understand the question.
There must be more to it than you have given us.
There must be a mistake in "p and q are the centers of C1 and C2 respectively".
What more can you tell us about it?

**Ilusa** · February 19th 2008, 01:23 PM

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'

**Plato** · February 19th 2008, 01:39 PM

Originally Posted by Ilusa

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'

Let $\left( {r_p } \right)\,\& \,\left( {r_q } \right)$ be the radii of the circles centered at p & q respectively.
By completing squares we get $\left( {r_p } \right)^2 = g^2 + f^2 - c\;\,\& \,\,\left( {r_q } \right)^2 = \left( {g'} \right)^2 + \left( {f'} \right)^2 - c'$ .
From the given $\left[ {l\left( {\overline {pg} } \right)} \right]^2 = \left( {r_p } \right)^2 + \left( {r_q } \right)^2$
What points are P & q?
Can you finish?

**Ilusa** · February 19th 2008, 01:50 PM

sorry I dont' get this bit
$<br /> \left[ {l\left( {\overline {pg} } \right)} \right]^2<br />$

the points p and q are not given but looking at the diagram the radii of the two circles are equal.

**Plato** · February 19th 2008, 02:11 PM

Originally Posted by Ilusa

sorry I dont' get this bit
$<br /> \left[ {l\left( {\overline {pg} } \right)} \right]^2<br />$ the points p and q are not given but looking at the diagram the radii of the two circles are equal.

$\left[ {l\left( {\overline {pg} } \right)} \right]$ is the distance from p to q.
P=(-g,-f) & Q=(-g',-f') so what is that distance?

**Ilusa** · February 19th 2008, 02:15 PM

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

**Plato** · February 19th 2008, 02:54 PM

$\left[ {l\left( {\overline {pg} } \right)} \right]^2 = \left( {g' - g} \right)^2 + \left( {f' - f} \right)^2 = \left( {g^2 + f^2 - c} \right) + \left( {\left( {g'} \right)^2 + \left( {f'} \right)^2 - c'} \right)$

**Ilusa** · February 19th 2008, 03:05 PM

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?

**Ilusa** · February 19th 2008, 03:08 PM

oh sorry it was the distance formula squared sorry! Thank you so much again!

**Plato** · February 19th 2008, 03:12 PM

Originally Posted by Ilusa

where do you get (g'-g)^2 + (f'-f)^2 from?

Do you know the formula for the distance between two points?

**Ilusa** · February 19th 2008, 03:14 PM

yeah I do thanks, I understand it fully now thank you again!

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