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

Explanation

The two circles are `x^(2)+y^(2)-2ax+c^(2)=0` and `x^(2)+y^(2)-2by+c^(2)=0`.
Respective centres are `C_(1)(a,0)` and `C_(2)(0,b)`.
Respective radii are `r_(1)=sqrt(a^(2)-c^(2))` and `r_(2)=sqrt(b^(2)-c^(2))`
Since the two circles touch each other externally , we have
`C_(1)C_(2)=r_(1)+r_(2)`.
`implies sqrt(a^(2)+b^(2))=sqrt(a^(2)-c^(2))+sqrt(b^(2)-c^(2))`
`implies a^(2)+b^(2)=a^(2)-c^(2)+b^(2)-c^(2)+2sqrt(a^(2)-c^(2))sqrt(b^(2)-c^(2))`
`implies c^(2)=sqrt(a^(2)-c^(2))sqrt(b^(2)-c^(2))`
`implies c^(4)=a^(2)b^(2)-c^(2)(a^(2)+b^(2))+c^(4)`
`implies a^(2)b^(2)=c^(2)(a^(2)+b^(2))`
`implies (1)/(a^(2))+(1)/(b^(2))=(1)/(c^(2))`