Question

Two circles $x^2+y^2+2ax+c=0$ and $x^2+y^2+2by=0$ touch if

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

Answer 1

Answer: (B)

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

$s_1=x^2+y^2+2ax+c=0$

$c_1=(-a,0)r_1=\sqrt{a^2-c}$

$S_2=x^2+y^2+2by+c=0$

$c_2=(0,-b),r_2=\sqrt{a^2-c}$

Since the two order touch each other,

$r_1+r_2=c_1{c}_2$

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

Squaring the above , we get

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

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

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

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

$c^2=(a^2-c)(b^2-c)$

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

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

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

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

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

$\therefore$ Option $\left(b\right)$ is correct