Q89 / 170 · Manipal (MET) 2016 · Mathematics
The condition that the straight line $c x-b y+b^2=0$ may touch the circle $x^2+y^2=a x+b y$ is
- A$a b c=1$
- Correct answerB$a=c$
- C$b=a c$
- DNone of these
Solution
We have, $c x-b y+b^2=0$ is a tangent of circle
$x^2+y^2-a x-b y=0$
$\therefore$ Centre $=\left(\frac{a}{2}, \frac{b}{2}\right)$, Radius $(r)=\sqrt{\frac{a^2+b^2}{4}}$
Distance from centre to the line $c x-b y+b^2=0$,
$r=\left|\frac{\frac{a c}{2}-\frac{b^2}{2}+b^2}{\sqrt{b^2+c^2}}\right|$
$\Rightarrow \quad \sqrt{\frac{a^2+b^2}{4}}=\frac{a c+b^2}{\frac{2}{\sqrt{b^2+c^2}}}$
$\begin{aligned} & \Rightarrow \quad\left(b^2+c^2\right)\left(a^2+b^2\right)=\left(a c+b^2\right)^2 \\ & \Rightarrow \quad b^4+b^2\left(a^2+c^2\right)+a^2 c^2=b^4+2 b^2 a c+a^2 c^2\end{aligned}$
$\begin{array}{rr}\Rightarrow & a^2+c^2=2 a c \\ \Rightarrow & (a-c)^2=0 \\ \Rightarrow & a=c\end{array}$
Hence, the correct answer is option: 'B'.