If the circles #x^2+y^2+2ax+c^2=0# and #x^2+y^2+2by+c^2=0# touch externally prove that #1/(a^2)+1/(b^2)=1/(c^4)#
Given that
the equation of first circle
#x^2+y^2+2ax+c^2=0#
#=(x+a)^2+y^2+c^2-a^2=0#
#=(x+a)^2+y^2=(sqrt(a^2-c^2))^2#
So its center #C_1->(-a,0)# and radius #r_1=sqrt(a^2-c^2)#
and the equation of 2nd circle
#x^2+y^2+2by+c^2=0#
#=x^2+(y+b)^2+c^2-b^2=0#
#=x^2+(y+b)^2=(sqrt(b^2-c^2))^2#
So its center #C_2->(0,-b)# and radius #r_2=sqrt(b^2-c^2)#
As the two given circles touch externally, the distance between their centers #C_1C_2# will be equal to sum of their radii.
#C_1C_2=r_1+r_2#
#=>sqrt(a^2+b^2)=sqrt(a^2-c^2)+sqrt(b^2-c^2)#
Squaring both sides we get
#a^2+b^2=a^2-c^2 +b^2-c^2+2sqrt(a^2-c^2)sqrt(b^2-c^2)#
#=>2c^2=2sqrt(a^2-c^2)sqrt(b^2-c^2)#
#=>c^4=(a^2-c^2)(b^2-c^2)#
#=>c^4=a^2b^2-b^2c^2-a^2c^2+c^4#
#=>b^2c^2+a^2c^2=a^2b^2#
#=>1/a^2+1/b^2=1/c^2#
