Lets transform the equations of the circles:
#x^2-2hx+h^2+y^2-2ky+k^2-h^2-k^2+c=0#
#(x-h)^2+(y-k)^2=h^2+k^2-c#
#(x-h')^2+(y-k')^2=h'^2+k'^2-c#
The centers of the circles are #P(h,k)# and #Q(h',k')#. Let #R(m,n)# be the point of intersection. Then the slopes of the lines #PR# and #QR# are, respectively:
#s=(n-k)/(m-h)# and #s'=(n-k')/(m-h')#
Because we have orthogonal intersection:
#s*s'=-1#
#(n-k)/(m-h) * (n-k')/(m-h') = -1#
#(n-k)/(m-h)=-(m-h')/(n-k')#
#(n-k)/(m-h)+(m-h')/(n-k')=0#
#((n-k)(n-k')+(m-h)(m-h'))/((m-h)(n-k'))=0#
#(n-k)(n-k')+(m-h)(m-h')=0#
#n^2-n(k+k')+kk'+m^2-m(h+h')+hh'=0#
#2m^2+2n^2-2m(h+h')-2n(k+k')+2kk'+2hh'=0#
Let the above equation be #(1)#.
Point #R# is placed on both circles (as point of intersection), so we will insert its coordinates into the equations of the circles:
#m^2+n^2-2hm-2kn+c=0#
#m^2+n^2-2h'm-2k'n+c'=0#
If we add those two equations we get:
#2m^2+2n^2-2m(h+h')-2n(k+k')+c+c'=0 (2)#
Lets rewrite #(1)# and #(2)# and subtract them:
#2m^2+2n^2-2m(h+h')-2n(k+k')+2kk'+2hh'=0#
#2m^2+2n^2-2m(h+h')-2n(k+k')+c+c'=0#
#2kk'+2hh'-c-c'=0# or
#2hh'+2kk'=c+c'#
