If two circles #x^2+y^2+2ax+c^2=0# and #x^2+y^2+2by+c^2=0# touch each other externally then prove that #1/(a^2)+1/(b^2)=1/(c^4)# ?

Let the given circles be,

# S_1 : x^2+y^2+2ax+c^2=0#.

#:. (x+a)^2+y^2=(sqrt(a^2-c^2))^2, (a^2-c^2) gt 0#.

Clearly, the centre #C_1# and the radius #r_1# of #S_1# are,

#C_1=C_1(-a,0), and, r_1=sqrt(a^2-c^2)#.

Analogously for # S_2 : x^2+y^2+2by+c^2=0#, we have,

#C_2=C_2(0,-b), and, r_2=sqrt(b^2-c^2), (b^2-c^2) gt 0#.

Given that #S_1 and S_2# touch each other externally.

Evidently, #"distance "C_1C_2=r_1+r_2#.

Utilising #C_1, C_2, r_1, r_2#, we get, from the above eqn.,

#(0-(-a))^2+(-b-0)^2=(r_1+r_2)^2, or, #

# r_1^2+2r_1r_2+r_2^2=a^2+b^2#.

#:. (a^2-c^2)+2r_1r_2+(b^2-c^2)=a^2+b^2#.

#:. r_1r_2=c^2#.

#:. r_1^2r_2^2=c^4#.

#:. (a^2-c^2)(b^2-c^2)=c^4#.

#:. a^2b^2-a^2c^2-b^2c^2+c^4=c^4#.

#:. a^2b^2=a^2c^2+b^2c^2=(a^2+b^2)c^2#.

Dividing throughout by #a^2b^2c^2!=0" (Why?)"#, we conclude,

#1/c^2=(a^2+b^2)/(a^2b^2)=1/a^2+1/b^2#, as desired!

Spread the Joy of Maths!

The first circle is

#x^2+2ax+y^2=-c^2#

#x^2+2ax+a^2+y^2=a^2-c^2#

#(x-a)^2+y^2=a^2-c^2#

The center is #(a,0)# and the radius is #sqrt(a^2-c^2)#

The second circle is

#x^2+y^2+2by=-c^2#

#x^2+y^2+2by+b^2=b^2-c^2#

#x^2+(y+b^2=b^2-c^2#

The center is #(0,b)# and the radius is #sqrt(b^2-c^2)#

If the circles touch each other externally, then

#"distance between the centers "=" sum of the radii"#

#sqrt(a^2+b^2)=sqrt(a^2-c^2)+sqrt(b^2-c^2)#

Squaring both sides,

#(sqrt(a^2+b^2))^2=(sqrt(a^2-c^2)+sqrt(b^2-c^2))^2#

#a^2+b^2=a^2-c^2+b^2-c^2-2sqrt((a^2-c^2)(b^2-c^2))#

#c^2=-sqrt((a^2-c^2)(b^2-c^2))#

Squaring both sides

#c^4=a^2b^2-a^2c^2-b^2c^2+c^4#

#a^2b^2=a^2c^2+b^2c^2#

Dividing by #a^2b^2c^2#

#1/c^2=1/a^2+1/b^2#

Are you sure of the #1/c^4# ?