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!
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!
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'
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?
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