Question

The coordinate of a point $P$ on the circle $x^2+y^2-4x-6y+9=0$ such that $\angle POX$ is minimum, where $O$ is the origin and $OX$ is the axis, are

• A. $(\frac{36}{13},\frac{15}{13})$
• B. $(-\frac{36}{13},\frac{15}{13})$
• C. $(\frac{14}{27},\frac{12}{27})$
• D. None of these

Answer 1

The correct option is A $(\frac{36}{13},\frac{15}{13})$

Given circle is $x^2+y^2-4x-6y+9=0$ ...(1)

Its centre is $C(2,3)$ and radius is $2.$

Let $OP$ and $ON$ be the two tangents from $0$ to circle (1),

then $\angle POX$ will be minimum when $OP$ is tangent to the circle at $P.$

Let$\angle POX=\theta ,$ then $\angle LCP=\theta$

Now, $CP=2,OC=\sqrt{2^2+3^2}=\sqrt{13}$

$\therefore OP=\sqrt{{OC}^2-{CP}^2}=\sqrt{13-4}=3.$

$\because C\equiv (2,3),\therefore OL=2.$

From the figure, $OM=OL+LM=OL+HP$

$\therefore OP\cos \theta =2+2\sin \theta$ or $3\cos \theta =2+2\sin \theta$

$\Rightarrow 3=2\sec \theta +2\tan \theta$ or $3-2\tan \theta =2\sec \theta$

$\Rightarrow 9+4{\tan }^2\theta -12\tan \theta =4(1+{\tan }^2\theta )$

$\Rightarrow 5=12\tan \theta \Rightarrow \tan \theta =\frac{5}{12}$

$\therefore \cos \theta =\frac{12}{13}$ and $\sin \theta =\frac{5}{13}.$

$\therefore P\equiv (OP\cos \theta ,OP\sin \theta )$

i.e.$P\equiv (\frac{36}{13},\frac{15}{13}).$