Question

The line $\frac{1}{r}=A\cos \theta +B\sin \theta ,$ touches the circle $r=2a\cos \theta ,$ then

• A. $a^2{B}^2+2aA=1$
• B. $a^2{A}^2+2aB=1$
• C. $a^2{A}^2+2aA=1$
• D. $a^2{B}^2-2aA=1$

Answer 1

The correct option is A $a^2{B}^2+2aA=1$

$\frac{1}{r}=A\cos \theta +B\sin \theta ,r=2a\cos \theta ,$
$\therefore \frac{1}{2a\cos \theta }=A\cos \theta +B\sin \theta$
Dividing by $\cos \theta ,$ we get
$\frac{1}{2a{cos}^2\theta }=A+B\tan \theta$
$\Rightarrow \frac{1}{2a}(1+{\tan }^2\theta )=A+B\tan \theta$
$\Rightarrow {\tan }^2\theta -2aB\tan \theta +(1-2aA)=0$
Above being a quadratic in $\tan \theta$ gives us two values of $\theta ,$ but if the line is to touch the circle, then the two roots of the above equation should be equal, the condition for which is,
${(-2aB)}^2-\mathrm{4.1.}(1-2aA)=0$
$\Rightarrow a^2{B}^2+2aA=1$