Question

$S$ is circle having center at $(0,a)$ and radius $b(b<a)$. $A$ is a variable circle centered at $(\alpha ,0)$ and touching circle $S$, meets the $x-$ axis at $M$ and $N$. If a point $P$ on the $y-$ axis such that $\angle MPN$ is independent from $\alpha$, then

• A. $\angle MPN={\cos }^{-1}\left(\frac{b}{a}\right)$
• B. $P\equiv (0,\sqrt{a^2-b^2})$
• C. $\angle MPN=-{\cos }^{-1}\left(\frac{b}{a}\right)$
• D. $P\equiv (0,-\sqrt{a^2-b^2})$

Answer 1

Answer: (A), (B), (D)

$P\equiv (0,-\sqrt{a^2-b^2})$

Let the radius of the variable circle is $r$

Now,
$\sqrt{a^2+{\alpha }^2}=b+r$
Let the point $P(0,k)$
We know that,
$M(\alpha -r,0)$ and $N(\alpha +r,0)$
Slope of line,
$MP=\frac{k}{r-\alpha }$
$NP=-\frac{k}{r+\alpha }$
Assuming
$\angle MPN=\theta \tan \theta =\left|\frac{\frac{k}{r-\alpha }+\frac{k}{r+\alpha }}{1-(\frac{k}{r-\alpha }\cdot \frac{k}{r+\alpha })}\right|\Rightarrow \tan \theta =\left|\frac{2rk}{r^2-{\alpha }^2-k^2}\right|\Rightarrow \tan \theta =\left|\frac{2k(\sqrt{a^2+{\alpha }^2}-b)}{(\sqrt{a^2+{\alpha }^2}-b{)}^2-{\alpha }^2-k^2}\right|\Rightarrow \tan \theta =\left|\frac{k}{-b}\left(\frac{\sqrt{a^2+{\alpha }^2}-b}{\sqrt{a^2+{\alpha }^2}+\frac{k^2-a^2-b^2}{2b}}\right)\right|$
As $\theta$ is independent of $\alpha$,
$\frac{k^2-a^2-b^2}{2b}=-b\Rightarrow a^2+b^2-k^2=2b^2\Rightarrow k=\pm \sqrt{a^2-b^2}$

$\therefore P\equiv (0,\pm \sqrt{a^2-b^2)}$
Now,
$\tan \theta =\left|\frac{k}{-b}\right|.\Rightarrow \tan \theta =\frac{\sqrt{a^2-b^2}}{b}\Rightarrow \cos \theta =\frac{b}{a}\therefore \angle MPN=\theta ={\cos }^{-1}\left(\frac{b}{a}\right)$