Locus of point of intersection of tangents at $A$ and $B$

Question

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?

Answer 1

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.

Answer 2

A syntetic aproach:

Let $I,J$ the tangent points in $C_2$ from $P$. Then $PI=PJ$. Moreover, by Pitagoras applyed to $\triangle PIA$, we have that $PI$ is the same value for all $P\in C_1$. On the other hand, as $OI\perp PA$ and $PA$ is a chord of $C_1$, then $I$ is the midpoint of $PA$. Thus $PA=2PI=2PJ=PB$. So, $PA=PB$ and this common value is the same for all $P$. As $PO=OA$ for all values independent from $P$, all triangles $OAP$ are congruents. In particular $\angle APO$ is independent from $P$.

Next, note that $\angle HAO=90$, cause $HA$ is tangent to $C_1$. Then, $\angle AOH=90-\angle AHO$. On the other hand, $\angle POA=180-2\angle APO=180-\angle APB=180-\frac{1}{2}\angle AOB=180-\angle AOH$. Then $\angle AOH+\angle POA=90-\angle AHO+180-\angle AOH=270-(\angle AHO+\angle AOH)=270-90=180$. It proves that $H$, $O$ and $P$ are collinear.

Therefore $\angle AOH=\angle OAP+\angle OPA=2 \angle OPA$. But the last is independent of $P$. Thus, $\angle AOH$ not depends on $P$. So, all triangules $\triangle AOH$ are congruents. Therefore $OH$ is independent of $P$. That is, $OH$ has the same value for all $P$. Then, $H$ is in a circunferecence centered in $O$.

Finally, we have $\sin\alpha=1/3$. But $3/OH=\cos\beta=\cos 2\alpha=1-2\sin^2\alpha=1-2/9=7/9$. Then $OH=27/7$.

Conclusion: The focus searched is $x^2+y^2=(27/7)^2$

Answer 3

Let the point of intersection of the tangents to the circle $x^2+y^2=9$ be $Q(h,k)$. The equation of $AB$, i.e. chord of contact to the larger circle is then $hx+ky=9$. Now if the half angle between the tangents drawn from $P$ is $\alpha$ we have $\cos 2 \alpha = \dfrac{7}{9}$. Then distance of $AB$ from center is $3 \cos 2 \alpha = \dfrac{7}{3}$.

This distance is also obtained as $\dfrac{9}{\sqrt{h^2+k^2}}$

Thus $\dfrac{9}{\sqrt{h^2+k^2}}=\dfrac{7}{3}$ from which we obtain the locus as

$x^2+y^2 = \left(\dfrac{27}{7}\right)^2$