Q94 / 200 · Manipal (MET) 2013 · Mathematics
The circles whose equations are $x^2+y^2+c^2=2 a x$ and $x^2+y^2+c^2-2 b y=0$ will touch each other externally if
- A$\frac{1}{b^2}+\frac{1}{c^2}=\frac{1}{a^2}$
- B$\frac{1}{c^2}+\frac{1}{a^2}=\frac{1}{b^2}$
- Correct answerC$\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}$
- DNone of these
Solution
The two circles are
$x^2+y^2-2 a x+c^2=0$
and $x^2+y^2-2 b y+c^2=0$
centres : $C_1(a, 0) \quad C_2(0, b)$
radii : $r_1=\sqrt{a^2-c^2}, \quad r_2=\sqrt{b^2-c^2}$
Since, the two circles touch each other externally. therefore
$C_1 C_2=r_1+r_2$
$\Rightarrow \quad \sqrt{a^2+b^2}=\sqrt{a^2-c^2}+\sqrt{b^2-c^2}$
$\begin{aligned} \Rightarrow a^2+b^2=a^2-c^2+b^2 & -c^2 \\ & +2 \sqrt{a^2-c^2} \sqrt{b^2-c^2}\end{aligned}$
$\Rightarrow \quad c^4=a^2 b^2-c^2\left(a^2+b^2\right)+c^4$
$\begin{array}{ll}\Rightarrow & a^2 b^2=c^2\left(a^2+b^2\right) \\ \Rightarrow & \frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}\end{array}$
Hence, the correct answer is option: 'C'.