JEE Main 2026 (Online) 21st January Morning ShiftQuestion 1
Let $PQ$ and $MN$ be two straight lines touching the circle $x^2+y^2-4x-6y-3=0$ at the points $A$ and $B$ respectively. Let $O$ be the centre of the circle and $\angle AOB=\frac{\pi}{3}$. Then the locus of the point of intersection of the lines $PQ$ and $MN$ is :
Circle
Solution
Concept. The two tangents from $P$ meet the circle at $A,B$; $OP$ bisects $\angle AOB$ and $\angle OAP=90^\circ$. So $OP$ is a fixed length and $P$ moves on a concentric circle. Formulas used. (1) Radius $\perp$ tangent at the point of contact(2) In right $\triangle OAP$: $OP=\dfrac{OA}{\cos\angle AOP}$Step 1 - Circle. $x^2+y^2-4x-6y-3=0$: centre $O(2,3)$, $r=\sqrt{4+9+3}=4$. Step 2 - Angle. $OP$ bisects $\angle AOB=\frac\pi3$, so $\angle AOP=\frac\pi6$.Step 3 - Use (1), (2). $OP=\frac{4}{\cos(\pi/6)}=\frac{8}{\sqrt3}$, so $OP^2=\frac{64}{3}$. Step 4 - Locus. $$(x-2)^2+(y-3)^2=\frac{64}{3}\Rightarrow 3(x^2+y^2)-12x-18y+39=64$$ $$\Rightarrow 3(x^2+y^2)-12x-18y-25=0.$$ Answer: C