# 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)?

Nov 25, 2017

If the circles ${x}^{2} + {y}^{2} + 2 a x + {c}^{2} = 0$ and ${x}^{2} + {y}^{2} + 2 b y + {c}^{2} = 0$ touch externally prove that $\frac{1}{{a}^{2}} + \frac{1}{{b}^{2}} = \frac{1}{{c}^{4}}$

Given that

the equation of first circle

${x}^{2} + {y}^{2} + 2 a x + {c}^{2} = 0$

$= {\left(x + a\right)}^{2} + {y}^{2} + {c}^{2} - {a}^{2} = 0$

$= {\left(x + a\right)}^{2} + {y}^{2} = {\left(\sqrt{{a}^{2} - {c}^{2}}\right)}^{2}$

So its center ${C}_{1} \to \left(- a , 0\right)$ and radius ${r}_{1} = \sqrt{{a}^{2} - {c}^{2}}$

and the equation of 2nd circle

${x}^{2} + {y}^{2} + 2 b y + {c}^{2} = 0$

$= {x}^{2} + {\left(y + b\right)}^{2} + {c}^{2} - {b}^{2} = 0$

$= {x}^{2} + {\left(y + b\right)}^{2} = {\left(\sqrt{{b}^{2} - {c}^{2}}\right)}^{2}$

So its center ${C}_{2} \to \left(0 , - b\right)$ and radius ${r}_{2} = \sqrt{{b}^{2} - {c}^{2}}$

As the two given circles touch externally, the distance between their centers ${C}_{1} {C}_{2}$ will be equal to sum of their radii.

${C}_{1} {C}_{2} = {r}_{1} + {r}_{2}$

$\implies \sqrt{{a}^{2} + {b}^{2}} = \sqrt{{a}^{2} - {c}^{2}} + \sqrt{{b}^{2} - {c}^{2}}$

Squaring both sides we get

${a}^{2} + {b}^{2} = {a}^{2} - {c}^{2} + {b}^{2} - {c}^{2} + 2 \sqrt{{a}^{2} - {c}^{2}} \sqrt{{b}^{2} - {c}^{2}}$

$\implies 2 {c}^{2} = 2 \sqrt{{a}^{2} - {c}^{2}} \sqrt{{b}^{2} - {c}^{2}}$

$\implies {c}^{4} = \left({a}^{2} - {c}^{2}\right) \left({b}^{2} - {c}^{2}\right)$

$\implies {c}^{4} = {a}^{2} {b}^{2} - {b}^{2} {c}^{2} - {a}^{2} {c}^{2} + {c}^{4}$

$\implies {b}^{2} {c}^{2} + {a}^{2} {c}^{2} = {a}^{2} {b}^{2}$

$\implies \frac{1}{a} ^ 2 + \frac{1}{b} ^ 2 = \frac{1}{c} ^ 2$