Let the given circles be,
# S_1 : x^2+y^2-4x-10y-7=0, and,#
# S'_2 : 2x^2+2y^2-5x+3y+2=0#.
We rewrite #S'_2" as "S_2 : x^2+y^2-5/2x+3/2y=-1#.
Then, completing the square, #S_2# becomes,
# S_2 : (x-5/4)^2+(y+3/4)^2=(3sqrt2/4)^2#.
Similarly, # S_1 : (x-2)^2+(y-5)^2=(6)^2#.
We find that the Centres #C_i" of "S_i (i=1,2)# are,
#C_1=C_1(2,5), and, C_2=C_2(5/4,-3/4)#.
If #P(X,Y)# is any arbitrary point on the reqd. common
chord, say #l#, then #P in S_1, and, P in S_2#.
#P in S_1 rArr X^2+Y^2-4X-10Y-7=0......(ast^1)#.
#P in S_2 rArr X^2+Y^2-5/2X+3/2Y+1=0......(ast^2)#.
#:. (ast^2)-(ast^1) rArr 3/2X+23/2Y+8=0, or,#
# 3X+23Y+16=0...........................................(ast)#.
We note that #(ast)# is a linear eqn. in #RR^2#, and,
therefore, represents a line passing through the points of
intersection of #S_1 and S_2,# their common chord #l#.
Switching from #(X,Y)# to the conventional #(x,y)#,
#:. l : 3x+23y+16=0#.
Next, to show that #l bot C_1C_2#.
We have, # l :y=-3/23x -16/3= #, so its slope #m_1=-3/23#.
The slope #m_2" of "C_1C_2" is "{5-(-3/4)}/(2-5/4)=23/3#.
#because m_1*m_2=-1, :., l bot C_1C_2#.
Hence, the Proof.