Let PQ and MN be two straight lines touching the circle x^2+y^2-4x-6y-3=0 at the points A …

Question

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 $\angle AOB=\dfrac{\pi}{3}$. Then the locus of the point of intersection of the lines $PQ$ and $MN$ is:

Accepted Answer

Centre $(2,3)$, radius $r=\sqrt{4+9+3}=4$. For an external point $(h,k)$, tangent length $=\sqrt{h^2+k^2-4h-6k-3}$.

Since $OA\perp PA$ and $OA=OB=r$, triangle $OAP$ has $\angle AOP=\dfrac{\pi}{6}$ (half of $\angle AOB$) and $\angle APO=\dfrac{\pi}{3}$, so $\tan\dfrac{\pi}{3}=\dfrac{r}{\text{tangent length}}$:

$$\sqrt3=\frac{4}{\sqrt{h^2+k^2-4h-6k-3}} \;\Rightarrow\; 3(h^2+k^2-4h-6k-3)=16.$$

Replacing $(h,k)\to(x,y)$: $3(x^2+y^2)-12x-18y-9-16=0 \Rightarrow 3(x^2+y^2)-12x-18y-25=0$.

Solutions · 1