Tangents are drawn from any point on the circle $x^2+y^2=9$ to the circle $x^2+y^2=1$ which intersect the circle $x^2+y^2=9$ at point $P$ and $Q$. Then find locus of point of intersection of tangent at $P$ and $Q$
Let any point in circle $x^2+y^2=9$ be $(3\cos\theta,3\sin\theta)$.
Then equation of tangent drawn from $(3\cos\theta,3\sin\theta)$ to circle $x^2+y^2=1$ is
$y-3\sin\theta=m(x-3\cos\theta)\Longrightarrow mx-y+3\sin\theta-3m\cos\theta=0\cdots (1)$
Above line is tangent to the circle $(x-0)^2+(y-0)^2=1^2$
So we have $\displaystyle |\frac{m(0)-0+3\sin\theta-3m\cos\theta}{\sqrt{9\sin^2\theta+9m^2\cos^2\theta}}|=1$
$\displaystyle (3\sin\theta-3m\cos\theta)^2=9\sin^2\theta+9m^2\cos^2\theta$
$\displaystyle 18m\sin\theta\cos\theta=0\Longrightarrow 9m(\sin 2\theta)=0\Longrightarrow \theta=\frac{\pi}{4}$
So equation of tangent line is
$\displaystyle mx-y+\frac{3}{\sqrt{2}}-\frac{3m}{\sqrt{2}}=0$
I did not know how I solve further
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Let $P$ be a point on $x^2 + y^2 = 9 $, and let $T$ be the tangency point (one of two) of the tangent(s) drawn from $P$ to the circle $x^2 + y^2 = 1$. The central angle $POT$ is given by
$ \angle POT = \theta = \cos^{-1} \dfrac{1}{3} $
If you draw the tangents from $P$ and $Q$ to the circle $x^2 + y^2 =9 $, then you'll construct two congruent right triangles with hypotenuse along $OT$ with one leg equal to $3$ and makes an angle $\theta$ with the hypotenuse, therefore,
$ \cos(\theta) = \dfrac{3}{h} = \dfrac{1}{3} $
There $ h = 9 $
Hence the locus of the intersection of all such tangents is a circle with a radius of $9$. That is, the locus is the circle,
$ x^2 + y^2 = 81 $