# Thread: Maths exam tomorrow need help with a circle proof please

1. ## Maths 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!

2. 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?

3. 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'

4. 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 $\displaystyle \left( {r_p } \right)\,\& \,\left( {r_q } \right)$ 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?

5. 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.

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

7. oh thank you, well I wrote that distance as (r^2+r'^2)^1/2
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

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

9. 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?

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

11. 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?

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