<s>[INST] Locus of point of intersection of tangents at $A$ and $B$ From a Point $P$ on $C_1 \equiv x^2+y^2=9$ two tangents are drawn to $C_2 \equiv x^2+y^2=1$ which meets $C_1$  at $A$ and $B$. Find the Locus of point of intersection of tangents at $A$ and $B$ on $C_1$.
My Try: Let $P(3\cos\alpha, 3\sin\alpha)$, $A(3\cos\beta, 3\sin\beta)$ and $B(3cos\gamma, 3sin\gamma)$. Now $PA$ is chord of $C_1$ and tangent to $C_2$ say at point $M$. Since center of two circles is same, let it be $O$.
Now $OM$ is perpendicular to $PA$ at $M$ and since $PA$ is chord of $C_1$, $M$ is Mid Point of $PA$. So coordinates of $M$ is $M(\frac{3\cos\alpha+3\cos\beta}{2},\frac{3\sin\alpha+\sin\beta}{2})$
Now since $M$ lies on $C_2$ we have
$$\left(\frac{3\cos\alpha+3\cos\beta}{2}\right)^2+\left(\frac{3\sin\alpha+3\sin\beta}{2}\right)^2=1$$ Simplifying we get
$$\cos(\alpha-\beta)=\frac{-7}{9} $$ By similar analysis
$$\cos(\alpha-\gamma)=\frac{-7}{9} $$ which means 
$$(\alpha-\beta)=2\pi+(\alpha-\gamma) \implies \gamma-\beta=2\pi \tag{1}$$
Now Equations of tangent to $C_1$ at $A$ and $B$ are
$$x\cos\beta+y\sin\beta=3$$ and $$x\cos\gamma+y\sin\gamma=3$$ Their Point of Intersection is
$$(h,k) =\left(\frac{3\cos\left(\frac{\gamma+\beta}{2}\right)}{\cos\left(\frac{\gamma-\beta}{2}\right)},\frac{-3\sin\left(\frac{\gamma+\beta}{2}\right)}{\cos\left(\frac{\gamma-\beta}{2}\right)}\right)$$
Now $$h^2+k^2=\frac{18}{1+cos(\gamma-\beta)}$$ Now using $(1)$ i will get the locus as $$h^2+k^2=9$$ which is the same circle $C_1$. Can i know where i went wrong and any better approaches?
 [/INST] 
We know that the required locus is a circle centered at O(0, 0). We only need to determine it radius.
For a particular case, that circle will go through Q(?, 0) when point P is at (3, 0).
The value of $\beta$ can be found from the brown triangle.
In the green right angled triangle, one angle is $2 \beta$.
The required radius = OQ can then be found. 
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