Question

If the circle $x^2+y^2+2ax+c=0$ and $x^2+y^2+2by+c=0$ touch each other, then

• A. $\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c}$
• B. $\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}$
• C. $a+b=2c$
• D. $\frac{1}{a}+\frac{1}{b}=\frac{2}{c}$

Answer 1

The correct option is A $\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c}$

Equation of circle:

$x^2+y^2+2ax+c=0$

$\therefore g=a,f=0,c=c$

$r_1=\sqrt{a^2+0^2-c}$

$\Rightarrow r_1=\sqrt{a^2-c}$

$Centre,C_1=(-a,0)and{r}_1=\sqrt{a^2-c}$

Equation of circle:

$x^2+y^2+2by+c=0$

$\therefore g=0,f=b,c=c$

$Centre,C_2(0,-b)and{r}_2=\sqrt{b^2-c}$

$C_1{C}_2=\sqrt{(-a-0{)}^2+(0+b{)}^2}$

$\Rightarrow C_1{C}_2=\sqrt{a^2+b^2}$

Since both circle touches each other

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

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

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

$\Rightarrow -2c+2\sqrt{a^2-c}\sqrt{b^2-c}=0$

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

$\Rightarrow (a^2-c)(b^2-c)=c^2\left\{squaringofbothside\right\}$

$\Rightarrow a^2{b}^3-a^{2}c-b^{2}c+c^2=c^2$

$\Rightarrow a^2{b}^2-a^{2}c-b^{2}c=0$

$\Rightarrow c(a^2+b^2)=a^2{b}^2$

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

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

Hence, option (a) is correct.