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^2)#?

1 Answer
Nov 25, 2017

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#